Question: Prove that there are no positive integers x and y, such that 2(x² + xy + y²) is a perfect square.

Solution: Assume that there is some k² = 2(x² + xy + y²); where x, y and k are positive integers.

LHS = RHS --> Even number
So k² is divisible by 2.

If k is an odd number, then k² is an odd number.

So k is not an odd number.

So k is an even number

k² is divisible by 4.
k²/2 is an even number.

k²/2 = x² + xy + y²
So x and y are both even numbers (because if either is an odd number, LHS = odd number)

If such a positive solution exists, then there must be a value whereby k² is minimum. Let k² = 2(x² + xy + y²) be such a value.

Since x, y and k are even numbers, let x = 2p; y = 2q and k = 2r whereby p, q and r are positive integers.

(2r)²/2 = (2p)² + (2p)(2q) + (2q)²
2r² = 4p² + 4pq + 4q²
r² = 2p² + 2pq + 2q²
r² = 2(p² + pq + q²)

But we have assumed that k² is the minimum.
r² = k²/4 is smaller than k² which results in a contradiction.

Hence there are no positive integers x and y, such that 2(x² + xy + y²) is a perfect square.


N.B.: If you have no clue what I just did, I proved the question by contradiction. I assumed the opposite and found out that the results contradict each other. I've shown that if there exist some positive integers x and y which satisfy the equation, then x/2 and y/2 must also satisfy it. By using the same method, I prove that x/4 and y/4 must also satisfy it, etc. Eventually I will end up to an odd number or a fraction which does not satisfy the equation. An elegant way of proving by contradiction!