Diophantine Equation 2
The integer solutions to x 2 + x y + y 2 = x 2 y 2 are ordered pairs ( x 1 , y 1 ) , ( x 2 , y 2 ) , … , ( x n , y n ) .
Find i = 1 ∑ n ( x i + y i ) .
If ( a , b ) is a solution, so is ( b , a ) , and the sum (the answer) involves both of them, i.e. i = 1 ∑ n ( x i + y i ) = a + b + b + a + ⋯ .
The answer is 0.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
4 solutions
There are also the solutions ( 1 , − 1 ) , ( − 1 , 1 ) . You should be more rigorous with your proofs. The idea is great, though. There are no consecutive non-zero integers that when multiplied together create a perfect square. Therefore, x y = 0 or x y + 1 = 0 , and the LHS then, in both cases, becomes 0 . therefore, x + y = 0 . This leaves us with the system of equations ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ { x y = 0 x + y = 0 or { x y = − 1 x + y = 0 , which creates the solutions, which, after checking, work.
Ah yes: I'd mistakenly assumed that x and y were non-negative! I sort of wrote this solution in a bit of a rush...
brilliant!!!
We can solve this problem with the following observation. If (x,y) is a solution, (-x,-y) is also the solutions. The sum of these solutions are 0. If either x or y is zero, we find two solutions of the form (x,0),(-x,0) OR (0,y), (0,-y). In these cases, the sum also reduces to zero.
I'm going to show you a few other useful techniques (another great technique is shown in my solution to the problem I've linked to in this question).
First of all, there's this lemma: ∀ a ∈ R , a 2 ± a = 4 ( 2 a ± 1 ) 2 − 1
And also the fact that the only perfect squares that are far away from each other by 1 are 0 and 1 .
x 2 + x y + y 2 = x 2 y 2 ⟺ ( x + y ) 2 − x y = ( x y ) 2 ⟺ ( x + y ) 2 = ( x y ) 2 + x y ⟺ ( x + y ) 2 = 4 ( 2 x y + 1 ) 2 − 1 ⟺ ( 2 ( x + y ) ) 2 = ( 2 x y + 1 ) 2 − 1 ⟹ { 2 x y + 1 = ± 1 2 ( x + y ) = 0 ⟹ x = − y ⟹ − 2 x 2 + 1 = ± 1 ⟹ x 2 = 2 1 ∓ 1 ⟹ ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ x 1 = 0 ⟹ y 1 = 0 or x 2 = 1 ⟹ y 2 = − 1 or x 3 = − 1 ⟹ y 3 = 1 ⟹ answer = i = 1 ∑ 3 ( x i + y i ) = 0
After checking, all 3 solutions work.
I just found it simpler to do is a,b is a solution, then -a,-b is a solution, therefore representing every solution can be written as a+(-a)+b+(-b)=0. Is my logic flawed?
Log in to reply
Haha, that's how I did it. :P
my big mistake :/ you're right, the logic is correct and using it, the problem is trivial. Should've asked for, e.g., i = 1 ∑ n x i y i , which, although would lead to the same answer 0 , couldn't be found out by such logic. Can't change it now for the reasons of making it fair for everyone.
Express x in terms of y and observe that the only y that satisfies is y=1 for which x is not difined therefore there are no solutions and the sum is 0
The equation can be rearranged to: ( x + y ) 2 = x y ( x y + 1 ) .
This implies that x y ( x y + 1 ) is positive and a perfect square. The product of two consecutive natural numbers is never a square, so the trivial (0,0) is the only integer solution.
Therefore, the answer is 0.