Double Square Sum

Problem

Consider the equation, $x^2 + y^2 = 2z^2$; if $GCD(x,y) = 1$ then the solution is primitive.

For example, $7^2 + 17^2 = 2 \times 13^2$ is a primitive solution, whereas $2^2 + 14^2 = 2 \times 10^2$ is not a primitive solution.

Prove that inifinitely many primitive solutions exist.

Solution

The integral solutions to the equation $a^2 + b^2 = c^2$ are called a Pythagorean Triplet, and it is well known that infinitely many primitive solutions exist.

Let $a$, $b$, and $c$ be a set of primitive Pythagorean triplets.

\begin{align}\therefore (a + b)^2 + (a - b)^2 &= (a^2 + 2ab + b^2) + (a^2 - 2ab + b^2)\\&= 2(a^2 + b^2)\\&= 2c^2\end{align}

Hence there are infinitely many primitive solutions to the equation $x^2 + y^2 = 2z^2$.

What about the equation $x^2 + y^2 = 3^2$?
Can you generalise for $x^2 + y^2 = kz^2$?

Problem ID: 333 (19 Nov 2007)     Difficulty: 3 Star

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