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Sum Of Squares And Multiple Of Product

Problem

Given that $x$, $y$ are positive integers and $x \ne y$, consider the following two results:

$5^2 + 13^2 = 194, 5 \times 13 \times 3 = 195$
$11^2 + 41^2 = 1802, 11 \times 41 \times 4 = 1804$

We can see that in both cases that the sum of the squares of $x$ and $y$ are almost a multiple of their product.

Prove that $x^2 + y^2$ can never be multiple of $xy$.


Solution

We are considering solutions to the equation $x^2 + y^2 = kxy$, where $k$ is some positive integer.

Suppose that $GCD(x, y) = h$; let $x = hm$ and $y = hn$.

Therefore $h^2 m^2 + h^2 m^2 = k h^2 mn \implies m^2 + n^2 = kmn$, where $GCD(m, n) = 1$.

So without loss of generality we can consider the primitive $m^2 + n^2 = kmn$.

As $m \ne n$ they cannot both be equal to 1, so let us suppose that $m \gt 1$. In addition, $m$ must contain a prime factor that $n$ does not contain; call this prime factor, $p$.

Clearly $m^2$ and $kmn$ are divisible by $p$, but $n^2$ does not contain this factor. Hence no integer solution exists and we prove that $x^2 + y^2$ can never be multiple of $xy$.

If $x$, $y$, and $z$ are positive integers, does $x^2 + y^2 + z^2 = kxyz$ contain solutions?

Problem ID: 300 (02 Jan 2007)     Difficulty: 3 Star

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