A Quadratic Diophantine Equation | Math Olympiads

🤩 Hello everyone, I'm very excited to bring you a new channel (aplusbi)
Enjoy...and thank you for your support!!! 🧡🥰🎉🥳🧡

If you need to post a picture of your solution or idea:
#radicals #radicalequations #algebra #calculus #differentialequations #polynomials #prealgebra #polynomialequations #numbertheory #diophantineequations #comparingnumbers #trigonometry #trigonometricequations #complexnumbers #math #mathcompetition #olympiad #matholympiad #mathematics #sybermath #aplusbi #shortsofsyber #iit #iitjee #iitjeepreparation #iitjeemaths #exponentialequations #exponents #exponential #exponent #systemsofequations #systems
#functionalequations #functions #function #maths #counting #sequencesandseries #sequence
via @YouTube @Apple @Desmos @NotabilityApp @googledocs @canva

PLAYLISTS 🎵 :

[moeberry]

The better way of solving this is to realize once you have the expression under the radical in the quadratic in x, we must have an upper bound on y, in this case 3y^2<or equal to 304. Which means y has to be at most 10 and at least -10 but by using symmetry you can bring it down to 11 values to test brute force. From 0-10 and not all will be perfect squares but a few are.

[factorial]

Because x and y is the same parity
we can write x +y = 2a, x – y = 2b
so x = a+b, y = a -b (for some integers a, b)
76 = x^2 – xy+y^2 = (a+b)^2 – (a+b)(a-b)+(a-b)^2 = a^2 + 3b^2
a^2 = 76 –3 b^2.
because 76 – 3b^2 is perfect square. b^2 = {4, 9, 25}
because of that a^2 = {64, 49, 1}
So x+y = 2a = {+-2, +-14, +-16}
One can check that (x, y) = (6, -4), (-6, 4), (10, 4), (-10, -4), (10, 6), (-10, -6) are valid solutions

[ibrahimmassy]

Very nice approach. I got x+y=+-2, +-14 and +-16. You can express x^2-x*y+y^2 as (x+y)^2/4+3*(x-y)^2/4, so 76 is expressed as the sum of an square and three times other square. The first square is related directly with x+y and give those solutions

[error__brain_deleted]

Problem is symmetric so assume y=x+c where x≥0 and c≥0 is an integer. Interchanging x and y does not change x+y.
x^2–x^2–cx+x^2+2cx+c^2–76=0; x^2+cx+c^2–76=0; x=[–c±√(304–3c^2)]/2;
304–3c^2≥0 must be a perfect square. c=4, c=6, and c=10 yield perfect squares.
Thus, x+y=±16, ±14, or ±2

[klausg]

You can reduce the problem very much by remarking that both x and y have to be even. Then you get a simpler problem
x^2-xy-y^2=304/4=76. Then again x and y have to be even. So you get again a simpler problem x^2-xy-y^2=76/4=19 and then solve the quadratic with very few solutions, easy to view at once.

[chaosredefined]

x^2 - xy + y^2 = 76

First, suppose x is odd. Then y^2 - xy = 76 - x^2. The RHS is odd, so the LHS is odd. Therefore, y^2 - xy is odd, and therefore y(y-x) is odd. The only way for a product of two integers to be odd is if both are odd, so if x is odd, then so are both y and y-x. But if x and y are both odd, then y-x is even. So, there is no solution when x is odd, and therefore all solutions will have an even value for x. By a symmetric argument, we also know that y will be even.

So, there exists integers a and b such that x = 2a and y = 2b. Substituting those in, we get 4a^2 - 4ab + 4b^2 = 76. Therefore, a^2 - ab + b^2 = 19. Also, note that if (a, b) = (m, n) is a solution, then (a, b) = (n, m) is a solution. Also, note that there is no integer n such that (a, b) = (n, n) or (n, -n). Therefore, without loss of generality, we can therefore assume that |a| < |b|. Therefore, a^2 < |ab| < b^2.

Now, as you went to do, we can see that a^2 - 2ab + b^2 + ab = 19, and therefore (a-b)^2 = 19-ab. Also, we can see that a^2 + 2ab + b^2 - 3ab = 19, so (a+b)^2 = 19 + 3ab. The nice thing about those equations is that (a-b)^2 and (a+b)^2 are both at least 1 (since (n, n) and (n, -n) are not solutions, we can't have a+b or a-b being 0). So, 1 <= 19 - ab, and 1 <= 19 + 3ab. Therefore -6 <= ab <= 18. Since |a| < |b|, this means that |a| has to be less than 4 (If |a| was 4 or larger, then |b| would be 5 or larger, and |ab| would be 20 or larger, but it can't be bigger than 18). So a = -3, -2, -1, 0, 1, 2 or 3.

By case work, we get:
a = -3 => 9 + 3b + b^2 = 19, so b^2 + 3b = 10. This occurs when b=2 or b=-5. Since we want |a| < |b|, we reject the 2 and get the solution (a, b) = (-3, -5)
a = -2 => 4 + 2b + b^2 = 19, so b^2 + 2b = 15. This occurs when b=3 or b=-5. This gives us the solutions (a, b) = (-2, 3) and (-2, -5)
a = -1 => 1 + b + b^2 = 19, so b^2 + b = 18. This has no integer solutions.
a = 0 => b^2 = 19. This has no integer solutions.
a = 1 => 1 - b + b^2 = 19, so b^2 - b = 18. This has no integer solutions.
a = 2 => 4 - 2b + b^2 = 19, so b^2 - 2b = 15. This occurs when b = -3 or b = 5. This gives us the solutions (a, b) = (2, -3) and (2, 5)
a = 3 => 9 - 3b + b^2 = 19, so b^2 - 3b = 10. This occurs when b=-2 or b=5. Since we want |a| < |b|, we reject the -2 and get the solution (a, b) = (3, 5)

So, our solutions are (a, b) = (-3, -5), (-2, 3), (-2, -5), (2, -3), (2, 5), or (3, 5). Therefore, (x, y) = (-6, -10), (-4, 6), (-4, -10), (4, -6), (4, 10), or (6, 10).

[SrisailamNavuluri]

If x, y are odd, lhs is odd.
If x, y are odd, even lhs is odd.
So both are even
Let x=2a, y=2b
4a^2-4ab+4b^2=76=4×19
a^2-ab+b^2=(a-b)^2+ab=19
2^2+5×3=19
(a, b)=(5, 3) or (-5, -3)
3^2+5×2=19.(a, b)=(5, 2) or (-5, -2)
19=25-6=(3+2)^2+(3×(-2))
( a, b)=(3, -2) or (2, -3)
Double the values and get x, y.
x+y=+/-16, +/-14, +/-2

[misterdubity]

I did (x-y)^2 + xy - 76 = 0 --> xy = 76 - (x-y)^2 and it wasn't any more difficult or time consuming than the presented method, except needed to check 8 different values of (x-y) manually.

[tixanthrope]

x^2 + y^2 >= 2xy
we can easily see x is not equal to y
so, x^2 + y^2 < 152 and x, y can be at most 12

we can easily see that x, y are both even

for x < y, we have y^2 > 76 and x^2 < 76, so x <=8 and y >= 10

by brute forcing the rest we can reach at the solutions.

[AzharLatif-dz]

This is absurd to cross add ( 6, 10) and (4, 10).How come you arrive at these coordinates? Then you concoct (16, 14) for X, Y values.This fails to satisfy x^2-xy+y^2=76.

[scottleung]

That's interesting - I actually got four solutions for x+y: +-2 and +-16.