I did it by considering that xy - 7 > x + y > sqrt(x^2+y^2) for all x,y >= 4. Therefore, if a solution exists, at least one of x,y must be less than 4. Direct calculation of the remaining cases revealed 0,7 7,0 3,4 4,3.
@MathCore-d8m5 ай бұрын
The Pythagorean triple part was very well thought out.
@محمدنیماکاظمی2 жыл бұрын
We can also use the inequality :(xy-7)
@niloneto16082 жыл бұрын
How can you be so sure that x+y>=xy-7?
@motherisape2 жыл бұрын
@@niloneto1608 that inequality is wrong
@D.Longley2 жыл бұрын
@@motherisape no it’s not. They just flipped the sides
@محمدنیماکاظمی2 жыл бұрын
For all integers x and y bigger than 1 xy is bigger than x+y .in this question x²+y²=(xy-7)² and we know that x²+y² is less than or equal to (x+y)² then you take sqrt of both xy-7 squared and x+y squared and for all cases that xy is bigger than 7 we get (xy-7)
@محمدنیماکاظمی2 жыл бұрын
Please think before you judge... I proved it in the comments replied to @Nilo Neto
@pranavmisra1552 жыл бұрын
What a channel this is!! I absolutely love your content and wish I could be as good as you in this algebraic gymnastics ☺️😊
@mathcanbeeasy2 жыл бұрын
If we write this in a quadratic equations in X, the discriminant must be a perfect square and the result will be much easy as P^2-Y^2=48. After that is very easy. Divisors of 48...
@johns.82462 жыл бұрын
This inspired my own similar problem. What are the integer solutions to x^4 - 9x^2 - 2xy - y^2 = 73 ?
@mcwulf252 жыл бұрын
I used the first approach. After some initial dead ends with differences of other squares. The Pythagorean approach is interesting.
@thepolishbear2 жыл бұрын
I know this could be solved algebraically, as he showed us, but I thought of differential equations to help solve this. His second solution used an introduction of two variables "a" and "b" to solve for "x" and "y" but I would have introduced implicit differentiation and differential equations from calculus. The squares would reduce from second degree to first degree terms, while instead of variables a,b I would have dy/dx ( or y') and dx/dy (or x') and the constant 7 would disappear, since derivatives are essentially about finding slopes of curves. But, I'm sure that maybe 🤔 only algebra should have been used for this to enhance algebraically thinking skills and advanced logical thinking on algebra and geometry terms.
@elkincampos38042 жыл бұрын
Note that (x,y) is solution if only if (y ,x) is solution, if only if (-x,-y) is solution. Then we can assume x>=y and x>0.
@metehan91852 жыл бұрын
0:44 why do we add +2xy ?
@Krish-i7q6 ай бұрын
To make x2 +y2 as a perfect square as (x+y)²
@faisalal-faisal14702 жыл бұрын
I initially tried the Pythagorean triples approach but then hit a stumbling block that you glossed over. The quoted formula giving (x,y,xy-7) in terms of (a,b) only generates primitive triples, which (x,y,xy-7) is not guaranteed to be (because they can share a factor of 7, for example). So your solution here is a bit incomplete unfortunately. To deal with the remaining case where say x=7k, y=7l, we're led to consider the equation k^2+l^2=(7kl-1)^2. I gave up at this point because this was starting to get a bit too long. Is there a quick way to salvage the Pythagorean triples approach?
@crustyoldfart2 жыл бұрын
" Is there a quick way to salvage the Pythagorean triples approach? " For a grubby and unrepentant pragmatist like me - YES. By inspection, start with either x =3, y = 4 or vice versa. then x*y = 12 and 12-7 = 5, giving the most common and best known PT. Oh and BTW, I personally prefer geometric solutions to apparently algebraic problems, for the simple reason that they show a real-world meaning as opposed to the purely abstract.
@nasrullahhusnan2289 Жыл бұрын
Using Pythagorean triplet a²+b²=c², with a=3 and b=4 then c=5. If x=3 and y=4, xy=12 --> xy-7=5, giving x²+y²=(xy-7)². Thus (x,y)=(3,4)
@aviraljain4652 жыл бұрын
Can you explain the Pythagoras part?
@harrysakata30822 жыл бұрын
I believe the Pythagorean part is wrong. I believe he is saying that since (x, y, xy - 7) is a Pythagorean triple, there exists a pair of integers (a, b) such that (x, y, xy - 7) = (a^2 - b^2, 2ab, a^2 + b^2). Euclid’s formula for generating Pythagorean triples states that (a^2 - b^2, 2ab, a^2 + b^2) is always a Pythagorean triple but does not generate all of them (Wikipedia says that (9, 12, 15) can’t be generated). I believe he is able to get the answer this way because the answers happened to be cases that the Euclid’s formula does not miss.
@somebody92325 ай бұрын
You should problem 4 from the same year, very fun!
@hinguyen79752 жыл бұрын
Could you explain the Pythagoras part? Which knowleadge does it use? Why exists a,b as integers since that is a right triangle?
@thane92 жыл бұрын
Step 1. We know if given a right triangle then a^2 +b^2 = c^2. That's how the Pythagorean theorem is typically used. But it's also worth noting that If a^2 +b^2 = c^2 then there exists a triangle with a right angle such that C^2 is the hypotenuse. It's the converse of the Pythagorean theorem and also in Euclid's Elements. Step 2. We know that because such a triangle exists, and we're using integers only for a, b, c then those integers must be a Pythagorean triple. It's how a Pythagorean Triple is defined. And by Euclid's formula for generating Pythagorean triples (editing to add: googling that formula will explain far better than I can) we know there are these other integers m and n where m>n>0, a=m^2-n^2, b=2mn, and c=m^2+n^2. Step 3 is what is shown in the video, the combining of those established definitions to get to the set of equations he uses to solve for the 3,4 portion. Not exactly what I did but it works. The 0,7 pair is so immediately obvious that it makes the 3,4 look a lot more complicated but it's not, it's only when adding rigor to the solution that things get trickier than they need to be, imho. Hope that helped!
@harrysakata30822 жыл бұрын
Harry Sakata 0 seconds ago I believe the Pythagorean part is wrong. I believe he is saying that since (x, y, xy - 7) is a Pythagorean triple, there exists a pair of integers (a, b) such that (x, y, xy - 7) = (a^2 - b^2, 2ab, a^2 + b^2). Euclid’s formula for generating Pythagorean triples states that (a^2 - b^2, 2ab, a^2 + b^2) is always a Pythagorean triple but does not generate all of them (Wikipedia says that (9, 12, 15) can’t be generated). I believe he is able to get the answer this way because the answers happened to be cases that the Euclid’s formula does not miss.
@harrysakata30822 жыл бұрын
@@thane9 Euclid's formula does not claim that all Pythagorean triples can be expressed using Euclid's formula. I think he arrives at the answer this way because the answers just happened to be ones that Euclid does not miss.
@geneyoungdho2 жыл бұрын
I initially thought ellipse and reactangle hyperbola formula
@arnavdas31392 жыл бұрын
At the beginning of the video I think you said for negative integers x & y
@jefflittle89132 жыл бұрын
"non-negative" This is what people say when they want to allow 0 also.
@RishabhSharma102252 жыл бұрын
That was really good.
@wesleysuen41402 жыл бұрын
Why is the graph so cute? (Plot it!)
@satrajitghosh81622 жыл бұрын
(xy -7 -x)(xy -7 + x) = y^2 case I : xy -7 -x = 1 xy -7 + x = y^2 ie 2x = (y-1)(y+1) y -1 = 2 , y +1 = x y = 3 , x = 4
@trillionman21052 жыл бұрын
Good one, maybe slightly harder problems are better
@marlongrau2462 жыл бұрын
Great! I got it. :)
@jumblefumble2 жыл бұрын
11:00 i didn't understand a single thing lol but I think that assuming is kinda risky , maybe this time not but other times it's not the most accurate thing
@thetempest02 жыл бұрын
Yeah, I didn't understand the 2nd approach, but the first one does make sense
@elkincampos38042 жыл бұрын
Another solution. Note if x,y are even or x,y are odds Mod 4 doesn't work. Then we can assume that x is even and y is odd. Let r= x*y. If r=6 or 8 then x=0 and x^2=1, That doesn't work. Note that (x+y)^2=x^2+y^2+2*x*y=(r-7)^2+2*r. And (x-y)^2=x^2+y^2-2*x*y=(r-7)^2-2*r. Thus (r-7)^2+-2*r are square. If r>8 then (r-8)^2=(r-7+1)^2=(r-7)^2+2*(r-7)+1=(r-7+2)=(r-7)^2+4*(r-7)+4=(r-7)^2+4*r-28+4. Thus 24>=2*r. r=12)). Cases 0
@syedaltaaf04022 жыл бұрын
Take x=0 , y=7
@Nishi-wc3zu2 жыл бұрын
2+6=?
@averyinterestingpineapple60382 жыл бұрын
8
@samarthgautam2 жыл бұрын
26
@와우-m1y2 жыл бұрын
asnwer=4 my munber isit
@the_luckyluke2 жыл бұрын
Mate U lost 2xy in the right Side ???
@motherisape2 жыл бұрын
You don't know algebra
@D.Longley2 жыл бұрын
x^2+2xy+y^2 factors to be (x+y)^2. That’s were 2xy went
@Unidentifying2 жыл бұрын
😂😂
@shashanks6262 жыл бұрын
@@motherisape Explain to him like Dwight did instead of being an ass.
@offbeatsofficial2 жыл бұрын
@@motherisape he's white or american so you know why 😂
@brianli34932 жыл бұрын
x=7 y = 0
@auwa692 жыл бұрын
2+2 = 5
@saswataghoshlearning43618 ай бұрын
i have solved it in a better way
@قاسمشبرنگ2 жыл бұрын
ok
@benjaminkarazi9682 жыл бұрын
Hello, I thank you for your beautifully performed processing, solution, and video. Fifty-one years ago, I used to do the math and did integrals for my girlfriend to solve and find the answers to become like, "I love you, I am yours, the life is beautiful, you are gorgeous, ETC. For reasons of the universe's laws of physic, mathematics or mathematicians always mimic physic to achieve formulas or formats doing what physic and or logic does and is accurate. The equation in this video made me curious to solve it my way, and it gave me more answers than expected; surprisingly, the equation has many solutions: y=6.0e⁰, x=1.66282713421e-¹ x=6.0e⁰, y=2.23371728658e⁰ y=1.1e¹, x=1.64751790548e⁰ x=1.1e¹, y=1.64751790548e⁰ y=2.8e¹, x=1.25099758593e⁰ x=2.8e¹, y=1.25099758593e⁰ y=1.090e², x=1.642678947e⁰ x=1.090e². y=1.642678947e⁰ y=3.0e, x=1.0233338915e⁰ x=3.0e³, y=1.0233338915e⁰ y=7.0e⁰, x=0.0e⁰ x=7, y=2.04166666667e⁰ y=1.0e⁵, x=1.00007000005e⁰ x=1.0e⁵, y=1.00007000005e⁰ y=5.0e¹, x=4.6879240832e⁰ x=5.0e¹, y=4.6879240832e⁰ y=1.36893e⁰, x=2.82216581448e⁰ x=2.82216581448e⁰, y=2.82216581448e⁰ y=0.0e⁰, x=7.0e⁰ x=0.0e⁰, y=7.0e⁰ y=2.063250108030.0e⁰, x=2.0e⁰ x=2.0e⁰, y=7.270083225280.0e⁰ y=3.0e⁰, x=2.0632501803.0e⁰ x=3.00.0e⁰, y=1.250.0e⁰ y=1.0e⁶, x=1.00000700000e⁶ x=1.0e⁶, y=1.00000700000e⁶ y=4.0e⁰, x=7.33333333333e-¹ x=4.0e⁰, y=2.999999999999e⁰ y=1.00000000000e-8, x=6.99999993000e⁶ x=1.00000000000e-8, y=6.99999993000e⁶ y=7.00000000000e-4, x=6.99510669260e⁰ x=7.00000000000e-4, y=6.99510339260e⁰ y=1.01897400249e⁰, x=3.39295773262e⁰ x=3.69000000000e², y=1.01897400249e⁰ ETC. For instance, the value of angles of the triabgle with x=3.69e², y=1.01897400249e⁰, and (xy-7.0e¹=3.69001406919e²) sides, are 9.0e¹°, 0.1582188808888e-¹°, and 8.98417811358e¹°; 9.0e¹°+0.1582188808888e-¹°+8.98417811358e¹°=1.80e²° Furthermore, I am neither trying to be an impertinent nor a rude person. Very best regards,
@EssentialsOfMath2 жыл бұрын
There are infinitely many pairs of x,y that satisfy the equation - but there are only four when you specify x,y to be non-negative integers, as in the problem.
@benjaminkarazi9682 жыл бұрын
@@EssentialsOfMath Hello, There are many more than four (4) nonnegative agents listed in the calculation. Regards,
@sanatanikonna2 жыл бұрын
Algebra is love ....thus guy is doing great have a look #mathmarvelasmr