If $a, b, c, d$ are natural numbers, such that, $ab = cd$, prove that $a^2 + b^2 + c^2 + d^2$ is a composite number.
It's not always divisible by $5$:
$6 \times 6 = 4 \times 9$ but $6^2+6^2+4^2+9^2 = 169$
If $a, b, c, d$ are natural numbers, such that, $ab = cd$, prove that $a^2 + b^2 + c^2 + d^2$ is a composite number.
It's not always divisible by $5$:
$6 \times 6 = 4 \times 9$ but $6^2+6^2+4^2+9^2 = 169$