Prove that if \(x,y,z\) are positive numbers satisfying \(x^2+y^2=z^2\), then \(x+y+z\) divides \(xyz\).

Solution

From the given condition:

\(x^2+y^2=z^2\)

We can rewrite it as:

\(x^2+y^2+2xy=z^2+2xy\)

\((x+y)^2=z^2+2xy\)

\(2xy=(x+y+z)(x+y-z)\)

We observe that \(x+y+z\) and \(x+y-z\) have the same parity, so \(x+y-z\) is divisible by \(2\).

Let:

\(x+y-z=2k\)

Then:

\(2xy=(x+y+z)2k\)

\(xy=k(x+y+z)\)

Multiplying both sides by \(z\), we get:

\(xyz=kz(x+y+z)\)

Therefore, \(xyz\) is divisible by \(x+y+z\).

Hence, the problem is proved.

Solution by: Thin Sokkean