
1. Let $\mathbb{N}$ be the set of all natural numbers and $S=\left\{(a, b, c, d) \in \mathbb{N}^{4}: a^{2}+b^{2}+c^{2}=d^{2}\right\}$. Find the largest positive integer $m$ such that $m$ divides $a b c d$ for all $(a, b, c, d) \in S$.


## Solution

Since $d^{2} \equiv 0,1(\bmod 4)$ at most one of $a, b, c$ is odd. Therefore 4 divides $a b c d$. Also, if 3 does not divide each of $a, b$ and $c$ then

$$
d^{2}=a^{2}+b^{2}+c^{2} \equiv 1+1+1 \equiv 0 \quad(\bmod 3)
$$

Thus 3 divides $a b c d$. Therefore 12 divides $a b c d$ and if $m$ is the largest positive integer such that $m$ divides all $a b c d \in S$ then $m=12 k$ for some positive integer $k$. But $(1,2,2,3) \in S$ and 1.2.2.3 $=12$. Hence $k=1$ and $m=12$.

## Remarks

The set $S$ is infinite because $\left(n, n+1, n(n+1), n^{2}+n+1\right) \in S$ for every positive integer $n$.
