Only 1 percent can solve this

Thanks to Ninad and Kulvardhan for the suggestion! The KZbin channel Vedantu Math shared a challenging problem saying only 1 percent can solve it. I admit I did not solve it. But I am happy to share the interesting solution. Special thanks this month to: Mike Robertson, Daniel Lewis, Robert Zarnke, Kyle, Lee Redden. Thanks to all supporters on Patreon! / mindyourdecisions
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[@MindYourDecisions]

For those curious, I recently posted a "shorts" video with a visual proof of the AM-GM inequality: kzbin.info0d59pTzdtaM?feature=share

[@demoncyborg7854]

When prash says only 1% can solve it he means it

[@skyfox.ew8h]

Hi,😅 Just wanted to know if it is acceptable to put it on the same base...16⁰= 1....same base is 16. Then I complete to the square for x and for y... Then, I'll get something like this (x+0.5)² + (y+0.5)² = ½ ??

[@leif1075]

That's a JOKE OR EXAGGERATIONS right?? Surely.mkst or many ppl.can see at least thst 1/4 is a solution and tnst it has to be less than 1??

[@leif1075]

But and y do NOT have to be the same .don't other ppl see thst is true??

[@gintokisan5171]

How did you get 0.25?? 3:28 Ok I get it now he distributed 0.5 (0.25+0.25) to x^2 + x then canceled 0.25 then the other 0.25 remained. Am I right??

[@tilakrupareliya3192]

When I saw the thumbnail, I immediately thought for "x=y" because of the symmetry of the equation. After putting x=y, you get x = y = -1/2 = -0.5 But I was not sure that it is the only solution. Good video 👍

[@kushagrapiano9036]

Same

[@AL_saoud_the_century_of_satan]

the symmetry of the equation does not mean x=y, it mean only S is a part of the solution of this equation f=(f)^-1 where f design the function f(x)=x²

[@Kartik_9512]

Any JEE Advanced aspirant can thought same bro😂

[@robertveith6383]

​@@Kartik_9512 * can think the same.

[@tilakrupareliya3192]

@@Kartik_9512 yup😁🤝🤝 We can answer a integration question by differentiating options 😂

[@deerh2o]

It's fascinating that to solve an equation, you start with an inequality.

[@divyanshkhanna8216]

I wanted to solve the question on my own, and I saw your comment without opening the comment section:)

[@mingmiao364]

Yes! And what you said is very true for real analysis, i.e., equality is often established through inequality. For example, to show two numbers are equal, we can do so by showing the difference between the two numbers is less than any positive number. Also, to show the limit equals some value, we use the “epsilon-delta” definition, which is an inequality statement at its core.

[@sheasewall7970]

@@mingmiao364 exactly what I thought of too, it's hard stuff

[@jessejordache1869]

I actually didn't get that whole thing. Turning 16 into 1/2 called for a conventional approach, which worked; the AM >= GM thing I've never seen before.

[@namphamnguyennhat2836]

Uh,well i solve this problem without using inequality?Am i wrong??

[@NestorAbad]

What an elegant solution! Thanks for sharing. I used a different approach to solve it, not so elegant but it worked: First, let (x,y) be a solution of the equation 16^(x^2+y)+16^(x+y^2)=1, then I look for the region in the plane where this solution must lie. To do this, note that both exponents x^2+y and x+y^2 must be negative. This is because 16^z>0 for all real z, and if z>=0, then 16^z>=1. So if one of the exponents was nonnegative, then 16^(x^2+y)+16^(x+y^2) >= 1 + positive > 1. So, the inequalities x^2+y= >= 1/2·(16^(-a+b) + 16^(a-b)) = = 1/2·(u + 1/u) where we made the substitution u=16^(-a+b). Now note that u>0, and it's easy to see that for u>0 the real function f(u) = u + 1/u reaches its absolute minimum at u=1, f(u)=2, so we can end our chain of operations writing 1/2·(u + 1/u) >= 1/2·2 = 1 and the >= is an = when u=1, that is, when 16^(-a+b)=1, so -a+b=0 and a=b. But in that case, if we recall our last expression before the first >= and put a=b we get: 1/2·(16^(-a+b+a^2) + 16^(a-b+b^2)) = 1/2·(16^(a^2) + 16^(a^2)) = 16^(a^2), and if we want 16^(a^2)=1 implies a=0, so a=b=0, as we wanted to show.

[@MindYourDecisions]

Wonderful! I had tried to work out something along these lines with no success. I like the approach because there is something symmetrical. One idea I was working on: now we know that x = y = -1/2 is the only solution, one might try to reverse engineer the solution. If one can prove that x = y, then it's straight-forward to see: 1 = 16^(x^2+y)+16^(x+y^2) = 16^(x^2+x)+16^(x+x^2) = 2 * 16^(x^2+x) = 2^(4x^2 + 4x + 1) This implies 4x^2 + 4x + 1 = (2x + 1)^2 = 0, giving the solution x = -1/2 so then y = x = -1/2. So it remains to prove x = y is necessary, and perhaps someone will work out the details :)

[@jayasmrmore3687]

@@MindYourDecisions one problem you can try making a video on is one I’ve tried to solve for many months: X/(y+z) + y/(x+z) + z/(x+y) = 4

[@kasiphia]

@@MindYourDecisions There is a much easier approach which I used. Initially I thought it wouldn't yield a full solution set when you brought out the AM-GM equality, but it indeed does. So first we make each power of 16 a power of 2, and then due to the symmetry of the problem, we can set each of the powers of 2 equal to 1/2, or 2 to the negative 1 power. This means we can create a system of equations where each of the powers of two are set to negative 1. Substituting y with its value in terms of x, yields the quartic equation 4x^4+2x^2+4x+5/4=0. And this equation has only one solution at x=-1/2. One could then substitute this value into the earlier derived expression of y and find that it is also -1/2. (Albeit it is a little tricky to solve the quarctic equation, which is why I left that out, but it's totally doable)

[@blake121666]

@@jayasmrmore3687 A linear transformation would turn that into abc = a + b + c where a = 2 * sqrt(2) * (y+z) b = 2 * sqrt(2) * (x+z) c = 2 * sqrt(2) * (x+y) a solution would be the first 3 integers because 1 * 2 * 3 = 1 + 2 + 3

[@HamishArb]

I couldn't prove it (although I didn't try for very long), but I did find the solution like so in about a minute. Notice that x and y are swappable (which is probably not the technical term lol, here meaning that if you swap x and y in the equation you get the same equation back, which also means that if you have x=1,y=2 as a solution for example, you immediately get x=2,y=1), therefore x=y is likely to be a solution, I checked if setting x=y makes sense and it was easy to find. Obviously this method doesn't guarantee every solution to be found nor does it guarantee any solutions at all, but it works quite often for these sorts of problems.

[@Ligatmarping]

Yes, that's nice. The technical expression you mentioned is like "since we are working with a symmetric function on x and y".

[@waheisel]

I also got the solution without coming close to proving it. I like "swappable"!🙂

[@paparmar]

So this is a very symmetric problem, as others have pointed out. Since the roles of x and y are interchangeable, we know that (x^2 + y) and (y^2 + x) are the same (note this does not necessarily mean x =y). This also implies that both LHS terms are equal (i.e., to 0.5 = 16^-0.25). At this point, we can infer all solutions to x and y must be negative fractions. Using 16 = 2^4, we then get the equations of two parabolas : one open-to-the-bottom vertical (y = -x^2 - 0.25) and one open-to-the-left horizontal (x = -y^2 - 0.25). Note the symmetry between x & y remains as it must. Now we can infer that there are at most four solutions, since such parabolas can intersect at zero, 1, 2, 3 or 4 points. It turns out these two parabolas have exactly one intersection, namely x = -0.5, y = -0.5. You can get the final solution by algebraic manipulations as detailed in this video, or by plotting (I used wolframalpha - but I wouldn't do that until I had distilled the original problem to this level). Personally, I like to focus on the properties of the potential solutions as much as I can before getting to their specific values.

[@picklesauce7983]

Note that x^2 + y + y^2 + x >= -1/2. Use Jensen's inequality to solve the rest after noting that the double derivative of 16^x is strictly positive. Equality case occurs at (-1/2,-1/2)

[@SyberMath]

Interesting! I made a video on this problem back in April 2022. I don't think it was a suggestion. I must have seen it in a book somewhere. Anyways, I think the way you write 16 as 4^2 at 2:27 and the rest of it with 0.5 being powers seems a little confusing for the average viewer. Overall a good video!

[@HeyWhySoSalty]

I thought the same. Seems like he assumes everybody will just know.

[@joanofarcsinus]

@@HeyWhySoSalty i don't know, i think anyone with decent highschool math could figure that out.

[@KavinUpreti]

where I saw it was in the "black book", a book used for joint entrance examination. maybe you saw it there too?

[@mohit12aExp]

Finally a challenging problem, I am glad you have gone back on maintaining difficulty standards on this channel. ( Caveat - I couldn't solve it)..

[@adamrak7560]

I call these problems "evil", because it was constructed to be punishing if you use your regular ways.

[@kousikthurlapati2927]

X+y=1 and (4*x^2-4*x+5=0) is also a possibility. But it gives complex values and still x=y holds good. The possible answers are x=y= (-1/2) or x=0.5+i and y=0.5-i Or x=0.5-i and y=0.5+i. As of now 3 sets of solutions

[@John-lf3xf]

The way I’d solve this is as follows. Took 2 mins. Since the equation is symmetric, we have that x=y for at least one solution. 2*16^(x^2+y)=1 16^(x^2+y)=1/2 taking log base 16 we have x^2+x=-.25 Then we solve this zero discriminate quadratic. There is only one zero here. x=y=-1/2

[@rakeshsrivastava1122]

Because of symmetry,each term should add up to 1/2.So x^2+y=-1/4.Similarly x+y^2=-1/4.Solving gives x=y=-1/2.

[@AniketKumar-lw6su]

I was able to solve it with a different approach, seeing the equations it is pretty clear that 4(x²+y)=4(y²+x)=-1 so that the equations will become 1/2 +1/2 To solve these equations i substituted one into another and ended up with a bi-quadratic equation 16x⁴+8x²+16x+5=0, to solve it i used the "Double cross method" and factored it as (4x²+4x+1)(4x²-4x+5), one real solution was given by the first factor which was x=-1/2 putting that in one the the first two equations i got y=-1/2 as well

[@allisthemoist2244]

These are so good for keeping the brain good at solving problems.

[@06nishant]

[16^(x+y-1)(x-y)]+1=16^(-x^2-y) Only possible when Either x+y=1 or x=y

[@zetacrucis681]

Exceptionally good problem and solution. So many in the comments missing the point that one needs to show that x = y (as was done in the video) not just state that it must be because I said so or because "symmetry of the problem". x + y = 1. Only solution must be x = y = 1/2, right? Cos symmetry, innit

[@idk-mw2tn]

what i did is just re-write 1 as 2*(16^-1/4) so then: x^2+y=-1/4 y^2+x=-1/4 and the equations are symmetrical so it is easy to see that x=y so then x^2+x=-1/4 so 4x^2+4x+1=0 then solve the quadratic to get (2x+1)^2 and x=y=-1/2

[@beausrsj]

Here's an algebraic way to solve it. Because 16^(-1/4)=1/2 we want to find _x, y_ so that x^2+y = -1/4 and x + y^2 = -1/4. If we subtract the two equations together, we get x^2 - x - y^2 + y = 0, which we factor into (x - y)(x + y - 1) = 0. Setting these equal to 0 yields x = y or x = 1-y. Start with the easier equation. x = y is satisfied when x and y are -1/2, and x^2+y = -1/4 and x + y^2 = -1/4. The result follows.

[@Cloud88Skywalker]

I just assumed by symmetry that both powers had to evaluate to 0.5 and thus bothh exponents should be -0.25. From there I solved the system of equations which is not particularly difficult. x^2 + y = -1/4 -> Isolate y -> y = -x^2 - 1/4 x + y^2 = -1/4 -> plug from above -> x + (-x^2 - 1/4)^2 = -1/4 -> develop the square and bring all to the left -> x^4 + (1/2)x^2 + x + 5/16 = 0 -> factor the polynomial -> (x^2 + x + 1/4) · (x^2 - x + 5/4) = 0 -> Solve for each factor: x^2 + x + 1/4 = 0 -> x = 0.5 = y x^2 - x + 5/4 = 0 -> x = 0.5 + i & 0.5 - i -> By symmetry y is the conjugate x. So, there you have the non-real solutions too.

[@ohplsshutup]

I have a much simpler solution LHS can be written as 2^(4(x^2+y)) + 2^(4(y^2+x)) This is equal to RHS when both powers are -1 (1/2 + 1/2=1) we get 2 equations for two variables and solve it!

[@mathcanbeeasy]

This is totally wrong. The ecuation is in R not in Q. As example, in R, the ecuation 16^a+16^b=1 have infinitely many solutions. You can have 16^a=0.12345, and 16^b=0.87655, with the solution a=log (base 16) (0.12345) b=log (base 16) (0.87655)

[@ohplsshutup]

@@mathcanbeeasy ah okay, I see it now, thanks for your reply!

[@twiedenfeld]

I rewrote as 16^xy(16^x+16^y)=1, then with some mental math figured it out it couldn't be positive, couldn't be zero, couldn't be -1 or anything smaller, so tried negative square and fourth roots until I found it.

[@johnnath4137]

If x > y, at least one of x^2 + y, x + y^2 is > 0 (there are 3 easy cases to examine, x > 0, x < 0, x = 0). This would entail the LHS of the given equation > 1. So x is not > y. Similarly, y is not > x. Thus we must have x = y = u (say). This leads to 1 =2 x 16^(u^2 + u) = 2 x 2^(4u^2 + 4u) or 4u^2 + 4u = 2^(-1) or 4u^2 + 4u + 1 = (2u + 1)^2 = 0 and u = x = y = -1/2.

[@shubhkapoor4057]

I was able to work my way through so as to say (with proof) that both x and y lie in the domain (-0.25,1). I was able to get the answer -0.5 in my head but was'nt able to furthur prove it by arithmetic.. good problem presh!

[@taflo1981]

I didn't see the rather elegant solution using the means inequality. My train of thought was way too complicated and went like this: First, I tried to find a solution with x=y, which means both exponents have to be -1/4 for both summands to be 1/2. Thus, one solution is x=y=-1/2. Then I looked at the function f(x,y) = x^2+y^2+x+y (i.e. the sum of the exponents). Since f(x,y) = (x+1/2)^2 + (y+1/2)^2 - 1/2, its value is strictly larger than -1/2 at every point other than x=y=-1/2. So for every such x,y, the larger of the two exponents would be strictly larger than -1/4, say z-1/4 (for some positive z), while the other exponent would be strictly larger than -z-1/4. This means that the original sum is strictly larger than 16^(z-1/4)+16^(-z-1/4) = (16^z+16^(-z))/2. Substituting u = 16^z > 1, we have that the original sum is strictly larger than (u+1/u)/2, which in turn is strictly larger than 1 for positive u (e.g. use derivatives to see that it is minimal at u=1).

[@phatphan1403]

My thought process is based on the exponent rule. We can not have 16^(something) + 16^(something) = 1 with a pair 0 + 1 or 1+0. It can also never be 2 - 1, -1 + 2. Then, the possibilities left of 16^(something) + 16^(something) = 1 must be a pair of 0.5 + 0.5, 0.75 + 0.25, 0.125 + 0.875 or something. Anything else in between will result in a very messy exponent of 16 that do not fit with the symmetry profile of x and y. I try to solve for x and y based on the first pair 0.5 + 0.5. Luckily, I got it on the first strike. x² + y = 0.25 x + y² = 0.25 => x = y = - 0.5

[@jeremiahlyleseditor437]

This completely confounded me. Thee GM whatever it was equality is completely new to me. I would have tried using Natural logs for the solution but you got it right.

[@aroundandround]

I was able to solve it entirely in my head as follows: Step 1: First, I guessed that x=y=-1/2 is a solution just by inspection. Step 2: We prove that there are no other solutions as follows. Suppose another solution was -1/2 + t and -1/2 + s. Then, the increase in the first exponent is t^-t+s and that better be matched by the decrease in the second exponent -(s^2-s+t), which yields t^2+s^2=0, so there is no other solution.

[@ccost]

i had simply figured out 16^-0.25 + 16^-0.25 = 1 and by using x=y because of the equations symmetry i had plugged x^2 + x = -0.25 into the quadratic formula and got x = -1/2

[@fried_ady6318]

There's also another solution, in the same case where x=y. The only thing that I thought was that 16^(x^2+y) and 16^(y^2+x) both equalled one half, so then their exponents would equal negative 1/4. So, since they were both hypothetically equal, I set x^2+y and y^2+x as equal, then grouped the x's and y's. This gave x(x-1)=y(y-1), and to find a standard solution, I proposed that y=x. Then, since y=x, x^2+x=1/4, thus x^2+x-1/4 equals 0, and through the quadratic formula, x and y both equalled the positive quotient of the sum of -1 plus root 2 all over two, giving 0.20710678118 for x and y.

[@samuelemallemi4706]

You left me spechless, that was a very simple and complete solution. Congratulation!

[@anandinilakantan2030]

Solution is so fascinating

[@aa01010]

4(x.x+y)=-1 and 4(y.y+x)=-1 gives us x=y . and that would help when x=y=-.5 hence the solution

[@TheWarmestWaffle]

My solution: Notice 16^n > 0 for all real n. Notice that for any real number pair x,y, we need to attain negative value for both x^2 + y and x + y^2 or else we will have either >1 + >0 > 0 or >0 + >1 >0. This implies negative restriction that each x,y must both be in (-1,0). This is because if not then at least one of x^2 + y and x + y^2 is guaranteed to be positive. Solution quantity by inspecting derivative form of powers in x,y: [2x and 1], [1 and 2y] implying that if 1 one solution, 1 > min(RHS) -> more than one solution min(RHS) occurs when equal symmetric powered positive bases have balanced negative powers so since they're symmetric min must be x,y=-0.5. Any other combination will be bigger than min(RHS). Since min(RHS) eval at x=-.5,y=-.5 results in LHS=RHS we have this as the only solution.

[@jarikosonen4079]

It should be seen that it is kind of symmetric equation and them maybe expect 1/2+1/2=1. But if its kind of diophantine, it makes hard to be solved.

[@theofficalblort2782]

i would just assume x and y are the same and that both must = 1/2. 1\2 = 16^(x+x^2), solve that and get -.5 all the same without all the extra stuff. I know assuming is bad but on a test thats the best course of action as you can guess and check

[@curingd]

Nice! I got as far as showing that x and y are between -1 and 0, then got most of the way there using the AM/GM inequality as a hint.

[@DrAndyShick]

I did this much faster, without doing pretty much any of this. Since they have the same base and nothing it said about the values of x, y, x^2, or y^2 specifically or relative to each other, and each exponent was in the same format (variable^2+ other variable), each exponent would be equal, and accordingly, each piece (16^to the polynomial exponent) would be equal to each other, specifically to equal 1/2. This would mean each exponent equaled -1/4 as 16^-1/4 =1/2. More specifically, for the reasons mentioned earlier, the value of x is indistinguishable from the value of y. Therefore, x^2+x=-1/4, or x^2+x+1/4=0. Using unFOILing or the quadratic, you get x=-1/2 (and same for y).

[@mssm9495]

There's a far, far easier way. 1) Notice that if x and y are swapped in the original equation: nothing changes. Therefore x=y. 2) Rewrite as (x^2+x)*ln(16)=ln(0.5) 3) Then solve using the quadratic formula.

[@lorenzograssi3637]

But the fact that they are swapped doesn't prove they are equal

[@jakob2946]

amazingly pretty derviation

[@antoine8278]

This was a very difficult problem.

[@greenpedal370]

I could probably have solved that when I was at college 43 years ago. Not a chance of doing it now!

[@gerger09]

My solution: 16^(x^2 + y) and 16^(x + y^2) both most likely equal 1/2 and 16^(-1/4) = 1/2. Since x^2 + y = x + y^2, x^2 - x = y^2 - y and we simplify to x = y. Because 16^(x^2 + x) = 16^(-1/4), (x is equal to y) we can reduce to x^2 + x = -1/4. We bring -1/4 to the other side, getting x^2 + x + 1/4 = 0. Factorizing, we get (x + 1/2)^2 = 0, and we can find that x = -1/2. x = y = -1/2 Were there any mistakes in my solution? I’m still in middle school so it’s likely I made a mistake somewhere. Great problem though!

[@keck24]

What a nice way to say 1*1=1

[@ericerpelding686]

At 2:37 I don't see how the second line became the third line.

[@dcterr1]

Geez, this one stumped me! Guess I'm not in the top percentile in solving these kinds of problems, aw well!

[@julianrenolfi5017]

I punt in a plot the surface z=... and then the surface z=1. The intersection was the same points you got. Not a resolution at all, but got to the solution.

[@CugnoBrasso]

I lightheartedly admit that I would have never been able to solve this, not in a million years.

[@pedrovelazquez138]

I found that both x and y have to be between -1 and 0. Like this: x E [-1,0) and y E [-1,0). The fact that both are equal to -0.5 is somehow cool... but I tried my best without the AM-GM property.

[@achintyavinod3322]

I wrote that 1 as (16)^-1/4 + (16)^-1/4... then x^2+y=y^2+x=-1/4... solved it to get the answer.

[@seanstotyn9267]

This was a fun one. I solved it without inequalities by first setting 4^(x^2+y) = sin u and 4^(x+y^2) = cos u since the equation then reduces to sin^2(u) + cos^2(u) = 1. Next I took partial derivatives with u(x,y) and exploiting the definitions in terms of trig functions: 1) (∂/∂x)sin u = cos u ∂u/∂x = (2x ln4)sin u 2) (∂/∂x)cos u = -sin u ∂u/∂x = (ln4)cos u 3) (∂/∂y)sin u = cos u ∂u/∂y = (ln4)sin u 4) (∂/∂y)cos u = -sin u ∂u/∂y = (2y ln4)cos u Taking the ratio of 1 and 2 to eliminate ∂u/∂x I found tan^2(u) = -1/2x and similarly for 3 and 4 I found tan^2(u) = -2y, so x and y must both be negative since tan^2(u) is positive. Equating both expressions for tan^2(u) leads to the condition that y = 1/4x. Now that u=u(x) I plugged in y = 1/4x into the sin u and cos u expressions and took total derivatives: 5) (d/dx)sin u = cos u du/dx = (2x - 1/4x^2)ln4 sin u 6) (d/dx)cos u = -sin u du/dx = (1 - 1/8x^3)ln4 cos u Taking (sin u)(5) + (cos u)(6) yields (after a bit of manipulation) 0 = (tan u sin^2(u) ln4)(1 - tan u)(1 - 8x^3), which has three solutions: a) sin u = 0 (not valid), b) x = 1/2 (not valid since x > 0), and c) tan u = 1. The only valid option is c) which leads to x^2 - x = y^2 - y, or plugging in y = 1/4x, the roots of the quartic function are +1/2 (order 3) and -1/2 (order 1), the only valid solution being x = -1/2 and y = -1/2.

[@bikwode]

why did I come up with two solutions, mine also contains -1/2 and -3/2. what did i get wrong?

[@alonbendavid7590]

I solves this problem by finding the minimum of f(x,y) of the left term being (-0.5,-0.5) and f of that is 1.

[@rasowa2958]

I cheated a bit and decided to search for solution where x=y so the equation should really be 1/2 + 1/2 = 1 Then it's easy: 16^(x^2 + x) = 1/2 2^4^(x^2 + x) = 1/2 2^(4x^2 + 4x) = 2^(-1) 4x^2 + 4x = -1 4x^2 + 4x + 1 = 0 (2x + 1)^2 = 0 2x + 1 = 0 x = -1/2 y = -1/2

[@kobenicat6795]

Wait you can simply just write 1 as 2^-1 + 2^-1. And turn the 16 to 2^4….. and you’ll have a system of equation

[@glam000]

Though no hard rule saying X can't be Y, but normally, anyone seeing both X and Y showing up in an equation simultaneously, a little voice would say to them while X means X, Y would mean something non-X. In the end, whatever real number or real numbers X and Y turn out to be, most people wouldn't jump to thinking that X also equals to Y, which is The Solution at the end of the clip. Conclusion: not only the math is hard, the language used to writing the question is just as insidious.

[@enockmuela7510]

Hi presh I tried the question using a different approach and realised that this question can only be simplified by completing the squares,therefore my answer came out as. (X+1/2)^2 +(Y+1/2)^2=2 And this cannot be simplified further

[@SenorQuichotte]

I used logs and solved by inspection

[@yiftachmaayan2249]

I bet everyone who solve this problem, assume from symmetry that x=y then 1/2 = 16^(x^2 + x) not a hard one

[@ronydey8826]

I need MCQ pattern answers for this type solution

[@demoncyborg7854]

When prash says only 1% can solve it he means it

[@incognitotorpedo42]

I am forever and always a member of The 99%.

[@pankajchavda6422]

I think there is no need of AM -GM THEOREM . As equation is symmetric x=y , it becomes symple

[@Misteribel]

x²+y²=1 is also symmetric, but has many solutions. The problem here, is proving that in this particular case, there is only one solution.

[@akhomeappliance3663]

Take log on both side

[@rohangeorge712]

i solved it the same way damn am-gm inequality is hella useful to know. proving that that the inequality must be an equality was hella satisfying for me and i was so proud of myself. then the rest was ez it was obvious the ONLY real solution is x=y=-0.5. which is not that hard to guess and check but it is satisfying to prove that thats the only real solution. pretty sure their are other ways to solve this without using am gm inequality so ima check the comment section. what i like about presh's problems is there are multiple creative solutions to solve the problem, making it more fun

[@kartikeyadimri]

This is question of advanced problems in maths for jee (aka black book )

[@aalvim27]

How about x=0 and y=-1/4?

[@hyeonsseungsseungi]

Awesome!

[@himanshukumar8881]

I solved it easily by assuming x=y 😊

[@farmertree8]

what a nice problem

[@ArthurShirinka]

This is a question from India's JEE mains exam

[@davidissel7980]

Couldn't you just "skip to the end" by rewriting the original equation with 16^0 = 1?

[@markus_park]

Wow. It’s actually so simple!

[@robtherub]

X squared plus y = -1/3 = x plus y squared. 16 to the minus third is a half right? So x minus 1/3 y 2/3?

[@ATOM-vv3xu]

yes that are results and i found them too but are those rly the only ones?

[@lorenzograssi3637]

I solved it but i couldn't prove it was the only possibile solution

[@louisrobitaille5810]

5:31 Goddammit... I literally got to the point of 0 = (x+0.5)^2 and 0 = (y+0.5)^2 but got too lazy 'cause I thought it didn't make sense and didn't wanna restart 🥲.

[@titan1235813]

Presh, my hat goes off to you because you're humble enough to admit, as a professional mathematician, that there are problems that you can't solve. Me neither, I couldn't solve this problem, especially come up with such a brilliant solution as using an inequality. Thank you!

[@mrmango2881]

Excellent problem. I solved using another method which is fairly elegant I would say :), though not as much as yours. It involves reduction to the parametric equation of a circle. No AM, GM inequality. Rewrote the equation as (4^(x^2+y))^2 + (4^(x+y^2))^2 =1 This is the equation of a circle A^2+B^2=1 in the variables A= 4^(x^2+y), B= 4^(x+y^2) Now I used the circle parametric equation in the parameter angle “t”. 4^(x^2+y) = sin(t) 4^(x+y^2) = cos(t) Observe that the LHS in both the equations is greater than zero, because 4^(some real number) > 0. So the RHS terms sin(t), cos(t) are both strictly positive. Apply log (to base 2 everywhere) to both equations. (We can do that because the terms are>0) x^2 + y = (1/2) log (sin(t)) x + y^2 = (1/2) log (cos(t)) Add both of the equations. x^2 + x + y^2 + y = (1/2) log (sin(t)cos(t)) Use the expansion of sin (2t) = 2 sin(t) cos(t) x^2 + x + y^2 + y = (1/2) log(sin2t) - (1/2)

[@mrmango2881]

Of course as a follow up, I’ll have to verify the solution by plugging x,y = -0.5 values into the equation.

[@mathcanbeeasy]

Excelent!

[@mr.nkundu5729]

Check your calculations bro......is it 1/2 log(sin2t) - 1/2log(2) ? And also give strong logic for why (something) -1/2

[@mathcanbeeasy]

@@mr.nkundu5729 log of something smaller than 1 is negative -1/2+ something negative is clearly less than -1/2. So, is correct what he did.

[@mrmango2881]

@@mr.nkundu5729 the log notation I used is to the base 2. So the (1/2) log 2 term which you correctly identified, becomes simply (1/2). As @MathCanBeEasy pointed out, -1/2 + something negative is less than -1/2. To understand why the term (1/2) log( sin 2t) is negative, observe that 0

[@rajinisekar6691]

I got the correct answer but my method of getting there is definitely wrong. I used this method when I figured I couldn't find 'all' real solutions, so decided to at least get one simple one 16^(x^2 + y) + 16^(x + y^2) = 1 If both terms were equal to 0.5, you'd have at least one of the possible answers, so, because 16 to the power of both terms is equal to 0.5: x^2 + y = -0.25 y^2 + x = -0.25 Rewrite them both to: x^2 + y + 0.25 = 0 y^2 + x + 0.25 = 0 putting together gives you x^2 + y + 0.25 = y^2 + x + 0.25 Which gives x = y Substituting x into y: x^2 + x + 0.25 = 0 That gives (x + 0.5)^2 = 0, So x = -0.5 and y = -0.5 Turns out to be the only right answer, but if I got this in my actual tests and there were more answers, I'd be fucked 😅

[@beebag6861]

Same, but i would do it because you get some points for at least one solution!

[@garvitmaheshwari6690]

I did the same

[@uzzionpathak2637]

You did everything right except didn't factor the equation (x^2 + y + 0.25 = y^2 + x +0.25) properly i think you solved it to get: x^2 - y^2 + y - x = 0 then solving further (x-y)(x+y) = (x - y) ------- eq-1(let) and saw that x = y makes both sides 0 and hence is a solution if not this at least something similar but what you should have done is taking (x-y) to LHS in eq-1 (x-y)(x+y) - (x-y) = 0 then take (x-y) as common factor (x-y)[x + y - 1] = 0 then you solve the two equations so obtained (x-y) = 0 and (x+y-1) = 0 basically swap x = y in the second equation and you'll get x = -1/2 = y That is how you can be sure this is the only right answer and no more roots exist.

[@poomonyoutube1607]

This is so beautiful. Simplifications like this is why I love algebra, the way something so complicated turns out to be something so simple... It's mesmerizing

[@Abedchess]

Without seeing the video Lemme guess, 🧐 0.5 + 0.5 =1, So 16^(x^2+y) =0.5 and 16^(x+y^2)= 0.5 x^2 +y = -1/4, x+y^2=-1/4 Hmm, x=y works as a condition So, if x = y, x^2 +x+1/4 = 0, x= -0.5, y=-0.5, Lol. That worked. So x= -0.5 and y= -0.5 is a solution

[@geoninja8971]

Yep, I'm one of the 99%.... :)

[@MindYourDecisions]

I read the solution and recalled it involved the AM-GM, but months later I was still not able to work it out. I think it's good to keep practicing. Even if you know the "trick", there is skill in applying it.

[@trendingthings1489]

@@MindYourDecisions yes you are right 😊 thank you for this motivation line & your videos are awesome, you are only one with this videos, keep uploading 😊.

[@venkatramana2298]

Let's add HM harmonic mean even..... It's going to effect a ton.... Thank you. From Venkat, India.

[@huzefa6421]

Me too 😂

[@willbishop1355]

Don't you have to plug x = y = -0.5 back into the equation to check that it's a solution? Using the AM-GM inequality proves that it's the only possible solution, but you still need to check that it works, unless I missed something.

[@Llortnerof]

No, since they were shown to actually be equalities rather than inequalities.

[@venkatramana2298]

As a math and physics faculty.... I appreciate you Presh.....god bless you..

[@AC130rus]

Bro what u doing in the night 10

[@veereshpt]

@@AC130rus what u doing in the night 10

[@theimmux3034]

mf two whole ass faculties

[@hssy2jrocker]

@@AC130rus what u doing in the night 10

[@deoxi]

@@AC130rus what u doing in the night 10

[@anuragmudgal5439]

thats a Regional Mathematics Olympiad(India) 2011 problem btw. Problem 6. pretty cool.

[@anamikadasgupta3644]

It's mentioned in Adv problems on mathematics by Vikas Gupta as well. Functions Q4

[@unknownredacted4028]

It never once occurred to me to use the A.M>/G.M>/H.M principle to solve this question

[@leif1075]

I don't think it would honestly.occur to ANYONE to use that right? Presh only used it because he had seen to before..but I don't see Ramanujan or anyone else using that..especially since thus doesn't explicitly involve geometry or averages

[@giovanniadelfio5378]

You could solve it in a much simpler way: x+y^2= -1/4(it comes from log of 1/2 with base 16).

[@MJ-we9vu]

The way to solve this equation is to make enough money to be able to hire a mathematician.

[@rjmunro]

It would have been good to put the numbers from the end in to the equation at the beginning to prove it works: x^2 + y = x + y^2 = (1/4) + (-1/2) = -1/4 16 ^ (-1/4) = 1/2 (1/2) + (1/2) = 1

[@Mr.Not_Sure]

Each time you see 2 unknowns in 1 equation, your first thought should be: let's rewrite it as an min/max problem.

[@jez2718]

Of note: if you allow x,y to be complex, then x = 0.5 + i y = 0.5 - i is also a solution.

[@crustyoldfart]

I would check that again !

[@kurzackd]

This was a nice solution, but I think you are missing one very important explanation, namely WHY this problem is actually so difficult: Namely, it is that the problem is to find ALL real numbers (not just 1-each for x and y) which satisfy the equation. FINDING the one -- x = y = -1/2 is EASY. PROVING that it is THE ONLY solution is NOT. :)

[@methodiconion8523]

I did this differently. I assumed x=y (figured while it may not necessarily be true, there is a solution such that it is.). 16^(x^2+x)=0.5 => x=-1/2 +- (1/2)sqrt(1-4ln(0.5)/ln(16))=-1/2

[@kespeth2]

I thought using logarithms at first, however IIRC the original + prevents it.

[@xekind]

If you know x^2+y = y^2 +x (which is actually the case), then you can use logarithms to simplify. Let's define p := x^2+y = y^2+x . Now we have 16^p + 16^p = 1 which is 2(16^p) = 1. Take the log of both sides and we have log(2) + p * log(16) = 0 . Simplify to p = - log(2) / log (16) which is p = - 1/4 or x^2+y = y^2 +x = -1/4

[@kespeth2]

@@xekind Nice.

[@rahevar3626]

Alternative solution: I think this way is also easier to understand 1 = 16^(x²+y) + 16^(y²+x) Notice that if we switch x and y the result will still be same therefore x=y 1 = 16^(x²+x) + 16^(x²+x) 1/2 = 16^(x²+x) 2^(-1) = 16^(x²+x) (2⁴)^(-1/4) = 16^(x²+x) 16^(-1/4) = 16^(x²+x) -1/4 = x²+x x² + x + 1/4 = 0 (x + 1/2)² = 0 x + 0.5 = 0 x = -0.5 y = -0.5 I solved it this way in my head but if there are any problems please let me know

[@florentinmunch6769]

The conclusion x=y just from symmetry is not working. Consider e.g. the equation x+y=0. This is also symmetric but has more solutions than x=y=0.

[@mageshts]

Is it possible to solve this problem using logrithmic method (considring base as 16 ) ?? Taking log on both sides of base 16

[@louisd95714]

This is exactly my thought, but I figured I am wrong because I am not great at Math.

[@Kartik_9512]

It's better understandable if we solve using logarithm.

[@mathcanbeeasy]

log(a+b) is not useful. Is not the same thing as log(a)+log(b). log(a)+log(b)=log(a*b)

[@mrmango2881]

Yes it is, I solved it in another comment. But before applying logarithm, I reduced the equation into two parametric equations (of circle).

[@mathcanbeeasy]

@@mrmango2881 i succed to made it with logs and no am-gm. I denote A=16^... And B=16^... So, A+B=1 and A*B=16^(x^2+y^2+x+y) Making log in base 2 in the second one and B=1-A: log(A)+log(1-A)=4(x^2+y^2+x+y) Not, the tricky part. f(x)= log(x)+log(1-x) have only one extreme point, which is x=1/2 and f(1/2)=-2. And from this we get 4*(x^2+y^2+x+y)

[@trnfncb11]

The expression is symmetric about x=y. So let x=u+d and y=u-d with d>0. You can immediately rewrite 16^(u^2+u+d^2)[16^a+16^(-a)], with a=d(2u-1). The expression in square brackets (like any z^t+z^(-t)) is >=2, while the factor outside has a minimum of 16^(-1/4)16^(d^2) = 16^(d^2)/2 for u=-1/2. So the expression is always >=1, with equality holding only if d=0, i.e. x=y=u=-1/2.

[@matthewfeig5624]

Great solution! Rewriting in terms of the sum and difference (like your u and d) is my go-to approach when given a symmetric expression.

[@mikhsly2]

Ohh

[@erikev]

Another way to solve this is to first just find a solution. Exponents must be negative, so try with x=y=-.25. Still too big. Try x=y=-1. Too small. Try x=y=-0.5. One solution found. Due to symmetry in x and y and the fact that x=y=-0.5 gives that lowest possible value of the expression, it must be the only solution. So this is a carefully constructed expression to make x=y=-0.5 the lowest value the right hand side can be, and that value is 1. Any other values give a bigger result. This is also why the inequality collapses to an equality.

[@Fred-yq3fs]

A tough one! I quickly found it'd be nice to deal with x2+y2+x+y as it factors nicely as a circle expression (x+0.5)^2+(y+0.5)^2 - 1/2, but I could not find how to get to that. I also knew I'd like to tackle this with an inequality as -1/2 was obviously a solution sitting there as an extremum with x and y constrained between the y=-x^2 and y = +-sqrt(-x) for x and y between -1 and 0. I said to myself: how to go from this sum to a product... but never did think about the am/gm inequality! Bummer. You really have to have a level of mastery to think about it as a bridge between sum and product. Thanks for that one!

[@foodybee4730]

Log(sec x)=1/2x²+Ax⁴+Bx⁶+........... Then find A and B ????? Only 1% people can solve this problem