Solving x+y=(x-y)^2, a Diophantine Equation

[@davidfriday7498]

For anybody interested in an oversimplified form of the final answer for this, x and y are consecutive triangle numbers. This leads to a pretty cool geometric interpretation if you feel like drawing pretty pictures.

[@SyberMath]

can you share a link to a drawing?

[@MichaelRothwell1]

Method 1 gives two sets of parametric solutions: 1. ((k²+k)/2, (k²-k)/2), or (k(k+1)/2, k(k-1)/2) 2. ((k²-3k+2)/2, (k²-k)/2), or ((k-1)(k-2)/2, k(k-1)/2) However, Method 2 just gives one set (the first). In fact, the second set of solutions in method 1 is the same as the first. To see this, just replace k by -k+1 in the first set and you get the 2nd set. This is easy to see using the factorised form.

[@jarikosonen4079]

it is checked that the square root gives integer out, but not checked the quadratic result is integer (that it is divided by two is still integer).

[@Catilu]

All the solutions are triangular numbers, he did not mention that, but if you for instance write n²+n as n(n+1), then you notice that's most definitely even and therefore divisible by 2. The same follows for all the other numbers looking similar

[@tonyhaddad1394]

Yes suppose n is odd so n +1 is even And if n is even so the expression is even

[@Jack_Callcott_AU]

It's interesting to note at 7:40 that all those number are triangular numbers. Triangular numbers occur when 2 consecutive integers are multiplied together and divided by 2, and they equal the sum of the first n, n-1, n-2 integers respectively.

[@geoffreyparfitt7003]

Yes. Apart from (0,0) other solutions are pairs of consecutive triangular numbers.

[@Blaqjaqshellaq]

Notice that the positive solutions for either variable are triangular numbers!

[@satrajitghosh8162]

Writing u for x-y one gets u - 2y = u^2 or (u -1/2)^2 = 1 -2y or u = 1/2 + √(1-2y), 1/2 - √(1-2y)

[@ffggddss]

x + y = (x - y)²After fiddling around with that equation from the thumbnail, I came up with 3 solutions: (x,y) = (0,0), (1,0), (6,3). [Also note the symmetry in x vs y; swapping them preserves equality/inequality. So (0,1) and (3,6) are also solutions. And we need only seek solutions with x > y; (0,0) is the only solution with x = y.] To get a bigger picture, you could plot z = LHS and z = RHS, in (x,y,z)-space, and see where they intersect (which will be a curve in 3-space). Then try to find where intersections are integers. z = x + y will be a plane that intersects the z = 0 plane in the line y = -x, and slopes upward with gradient vector ∇z = (1,1). z = (x - y)² is a parabolic cylinder opening upward, with "trough line" y=x, z=0; i.e., (x, x, 0) ∀x. That's as far as I've gone for now. Let's see what else there is . . . Just now seeing the comment from michael purtell, and I think he's got it: • Choose any integer n. • (x, y) = (½n[n+1], ½n[n-1]) is a solution . . Note that these are both integers, b/c "n" or the other factor must be even. They are, in fact, just consecutive triangular numbers. Then LHS = x + y = n² ; x - y = n, so that RHS = (x - y)² = n² And as stated earlier, (y, x) will also be a solution. Those would appear to be all possible solutions. Hat-tip and kudos to michael purtell. Fred

[@SyberMath]

Thank you, Fred!

[@boguslawszostak1784]

It becomes easy when we use my favorite substitution. x = m + y y = m-y what after substitution to the equation x + y = (x-y) ^ 2 gives 2m = (2y) ^ 2 m = 2r ^ 2 from where we have the general solution directly x = 2r ^ 2 + r = r (2 * r + 1) y = 2r ^ 2-r = r (2 * r - 1) When we throw solutions in integers, x-y = 2r is also an integer so t = 2 * r r = t / 2 x = (t / 2) * (2 * t / 2 + 1) = t * (t + 1) / 2 y = (t / 2) * (2 * t / 2-1) = t * (t-1) / 2 this is the answer. checking we have one of the numbers t, t + 1 must be divisible by 2 so for any given t x is an integer one of the numbers t, t-1 must be divisible by 2 so for any given t, y is an integer x + y = t * (t + 1) / 2 + t * (t-1) / 2 = t * (t + 1 + t-1) / 2 = t ^ 2 x-y = t * (t + 1) / 2-t * (t-1) / 2 = t * (t + 1-t + 1) / 2 = t

[@tomctutor]

A few comments below refer to triangular numbers.✓ if we let x -y = n, then x+y = n+2y n +2y = n^2 y = (n^2-n)/2 = n(n-1)/2 for some integer n = ∑_(m=0,m=n-1) m = 0+1+2+...n-1 x = y+n = ∑_(m=0,m=n) m = 0+ 1 +2+ ...n or vica versa by symmetry of solution set (x,y) both x,y are pairs of successive triangular numbers; e.g. (0,0), (0,1), (1,3), (3,6) ... and so on

[@yoav613]

Nice,the second method blowed my mind!💥💥

[@SyberMath]

Glad to hear that!

[@kushaldey3003]

@@SyberMath you always save the better methods for later right? 😏

[@vladimirkaplun5774]

you should have noticed that the sum of 2 neighbor sums of the arithmetic progression is a full square. it is what n(n+1)/2 and n(n-1)/2 are

[@michaelpurtell4741]

Must have missed something. To me it seems that there are an infinite number of solutions pick any N sum the consecutive integers including N that is X subtract N that is Y ( or the same thing multiple any two consecutive number divide by 2 that is X subtract the lesser one that is Y)

[@mthtactics9195]

I challenge you to find x in A^x^x + Bx^x + C = R in terms of A,B,C

[@MgtowRubicon]

The solutions are pairs of the sum of consecutive integers. k(k-1)/2 and k(k+1)/2. The other "solution" is just a variant.

[@echandler]

Nice problem. Using method 2, replacing k with -k interchanges x and y. Done.

[@SuperYoonHo]

Hey? I saw this problem somewhere but you'r solution is REALLY cool! AWESOME video!😎👌

[@tanishkraj5140]

with the help of your videos i have developed interest in mathematics thanks

[@SyberMath]

That's great! 🥰

[@maxm9960]

Assume y>=x and y=x+a, a>=0. Then LHS=2x+a. If x0, then because x,y are integers x

[@MichaelRothwell1]

There is no need to assume a≥0. The solution works for all integer values of a.

[@MrLidless]

Without loss of generality, let x>= y, ie x = y + n, n>=0. 2y + n = n² 2y = n² - n = n(n - 1) y = n(n - 1)/2 x = n(n + 1)/2

[@robertveith6383]

Use "^" for exponentiation. So, write n^2 for n squared and t^2 for t squared.

[@rakenzarnsworld2]

Level 1: x = 1, y = 0 Level 2: x = 3, y = 1 Level 3: x = 6, y = 3 Level 4: x = 10, y = 6 Level 5: x = 15, y = 10 Level n: x = {n(n+1)}/2, y = {n(n-1)}/2

[@Yuenth-q3h]

Integer solution

[@angelmendez-rivera351]

Let m = x - y, hence m + 2·y = m^2, so 2·y = m^2 - m, and y = m·(m - 1)/2. Therefore, m = x - m·(m - 1)/2, which implies x = m·(m + 1)/2. From here, it is easy to prove that these satisfy the equation for all integers m. In fact, if y(m) = m·(m - 1)/2, then x(m) = y(m + 1), and every pair (x(m), y(m)) satisfies the equation.

[@SyberMath]

Nice!

[@a97304]

Zero is an integer. There are more complex answers, but this works if x=1 and y = 0. Figured this out in my head in a few seconds. The long answer is more 'interesting'. Other than just love of math puzzles, is there much in the way of practical use for this question.....other than brain exercise?

[@9WEAVER9]

ALL HAIL THE GRADIENT

[@HoSza1]

Hi, I was doodling on equations and I found this one: f(x)f(-x)=f(x^2). I can find one solution, because of how it was created, but I don't know, how to find all of them?

[@angelmendez-rivera351]

Two functions f : Q -> Q that solve this equation, are f(x) = 0 everywhere and f(x) = 1 everywhere. Every f that satisfies the equation satisfies the condition f(0)^2 = f(0), which implies f(0) = 0 or f(0) = 1 is a zero divisor. Additionally, f(1)·f(-1) = f(1), which means f(1) = 0, or f(-1) = 1. If f(1) = 0, then f(-1) can be any rational number, and if f(-1) = 1, then f(1) can be any rational number. Consider the set of positive rational numbers that are not the square of another rational number, K. To construct a solution f for the functional equation, you get to freely choose f(q) for every q in K. You also get to freely choose f(-q^(2^m)) for every q in K and every positive integer m. Once you have done this, f(q^2) will be uniquely determined for all q in K, and so will be f(q^(2^m)) for all q in K and all positive integers m. Every such choice is constitutes a new function that solves this functional equation, and in fact, this construction method covers every solution of the functional equation. The same solution method applies if you replace Q with any proper subring of the constructible numbers. Let L be the ring of constructible numbers. In this case, there actually are no square free elements. So instead, f(x^2) = f(-x)·f(x) is equivalent to f(x) = f(-sqrt(x))·f(sqrt(x)) = f(-x^(1/2))·f(x^(1/2)) = f(-x^(1/2))·f(-x^(1/4))·f(x^(1/4)) = f(-x^(1/2))·f(-x^(1/4))·f(-x^(1/8))·•••, and here, we have some problems. We need some topology in order to have a notion of sequence convergence, and then, we must partition Q into equivalence classes so that we have no issues with sequences interfering with each other in this product. A similar construction as the one from the previous paragraph will follow, and this also applies for non-Gaussian supersets of L, such as the real numbers. If we work with the ring L[X]/(X^2 + 1) instead, or a superset thereof, such as the algebraic numbers or the complex numbers, then it gets even more complicated, since now you must deal with roots of unity.

[@HoSza1]

To simplify things (and restrict the seemingly overly large set of solutions) I would add that the solution f must be a real valued function on real inputs and may have only a finite number of discontinuities.

[@angelmendez-rivera351]

@@HoSza1 I am not sure if any such solutions exist. Discontinuities seem to be fundsmentally built into the functional equation. Even if such solutions do exist, I do not know of a construction that guarantees them, or a proof that would prove their existence.

[@HoSza1]

@@angelmendez-rivera351 My first solution was f(x)=1/(1+x+x^2) but then I noticed f(x)=1+x+x^2 is a solution as well. Both of these are continuous for all real values.

[@HoSza1]

Interestingly wolframalpha does find a different solution that has one discontinuity at x=0: f(x)=-(1+x+x^2)/x

[@JPTaquari]

(X + Y ) + 2XY = X² + Y² 1) 0 and 0 2) 1 and 0 3) 1 and 3 4) 3 and 1 5) 6 and 3 6) 3 and 6 7) 0 and 1 Do you want more???

[@sunnyzhu5791]

what kind software do you use to make this video?

[@SyberMath]

Notability App

[@giuseppemalaguti435]

, y=1,,3,6,10,15,21

[@pageegap]

The solution begins with y=O then thé other values as you wrote...

[@MichaelJamesActually]

this is probably a dumb question. however, when you get to your set solution, ((n^2+n)/2, (n^2-n)/2), and the other one, how do you know that these will always be integers (given that n is an integer)?

[@kushaldey3003]

n²+n = n(n+1), which is a product of 2 consecutive integers, one of them must be even, implying their product is also even, same for the other one

[@antonyqueen6512]

It is not dumb. The video was missing this point. Factorising the solutions gives a product of two consecutive numbers for each of x and y, one of the consecutive numbers lust be even (and the other one odd) and the product is therefore an even number.

[@MichaelJamesActually]

@@kushaldey3003 thank you!

[@MichaelJamesActually]

@@antonyqueen6512 thanks as well.

[@antonyqueen6512]

@@MichaelJamesActually ☺️ I had actually posted a comment about this yesterday. Some liked and one was just rude🤗

[@starpawsy]

THere is the trivial solution x = y = 0 . THis should be mentioned at the start and then never mentioned again.

[@jmart474]

Cool. Thank you.

[@antonyqueen6512]

The solution is not complete. In the first method, you have still to find the integers n (limitations on n if any) such that y an x are integers since the result is written in fractions. y=(n^2-n)/2 => y=n(n-1)/2 Since one n or n-1 has to be even for any integer n => n(n-1) is even number => solution y=(n^2-n)/2, n being any integer. Same reasoning apply for x. I f you went the extra step to write y=(4n^2-4n)/8 instead of directly writing the result, the steps of showing y=(n^2-n)/2 is integer for any n is more important and should also be done 😉

[@pietergeerkens6324]

(n^2 + n ) / 2 = [ n (n+1) ] / 2 is the well known binomial coefficient n_C_2; and is always an integer.

[@antonyqueen6512]

@@pietergeerkens6324 as part of the solution it has to be said, mentioned !

[@pietergeerkens6324]

@@antonyqueen6512 Don't be absurd. This isn't a channel for 12 year olds.

[@stevenlitvintchouk3131]

@@pietergeerkens6324 12 years old is just about right, because my teacher proved that by induction when I was about that age.

[@johns.8246]

Hi. I'm trying to find values of positive integers x, y, and n, such that x^3-y^3=2^n . I really could use some help.

[@WahranRai]

x, y must be even and n >= 3 etc... An obvious solution : x=2, y=0, n=3 ---> x=2^a, y= 0 and n=3*a with a integer >=1

[@Yoonjh834]

why 8y+1=odd?

[@stevenlitvintchouk3131]

I did it with the first method.

[@GourangaPL]

1st method seems easier for me

[@bussi7859]

X=1 y=0 or x=0 y=0

[@beeruawana6662]

Very good question sir 🙏🙏. Thank you so much sir 🙏🙏

[@SyberMath]

Most welcome

[@jancermak1988]

x=6, y=3. Easy.

[@SyberMath]

Is that it?

[@jancermak1988]

@@SyberMath Yes

[@dandeleanu3648]

@@SyberMath let x+y = a and x-y = b then a=b^2 and a+b=2x and we get b^2+b-2x=0 from last quadratic equation we inferr that 1+8x = m^2 for m natural. m^2 - 1 =8x (m+1)(m-1)=8x m-1=x and m+1=8 m=7 and x=6 if x=6 then y= 3 or y=10 solution: (6;3) (3;6) (6;10) (10:6) the same for (1:3) (3;1) (1:0) (0;1)

[@dracokinerek]

x=0 y=0

[@marklevin3236]

Too complicated let x-y=z..... For any z we have a linear system x-y=z. x+y=z^2. For any given z there is an unique solution X=(z^2+z)/2 y=(z^2-z)/2

[@dog.3162]

literally a solution used in the video

[@williamhogrider4136]

Wow 🍺🍺🍻.

[@michaelyap939]

0 and 0

[@marshalls36]

好，奖 1 万元

[@actions-speak]

哈哈哈

[@barakathaider6333]

👍

[@bussi7859]

Make it complicated

[@tbg-brawlstars]

First!!!

[@kylecow1930]

ok so let z = x-y so now z^2 = z+2y, z^2 - z - 2y = 0, by rational root theorem z|2y so let 2y=az a some integer then z^2 - z + az = 0 so z(z-1+a)=0 so z = 1+a so as 2y=az, y=a(a+1)/2, subbing back to z=x-y, x=(a+1)+a(a+1)/2 = (a+1)(1+a/2) = (a+1)(a+2)/2 also the equation is symetric in xy so let f_n = n(n+1)/2, our solutions are (f_n, f_(n+1)), (f_(n+1), f_n) for any integers n