IRL ABooklet

Solution 1. First note that $2023 = 7 \cdot 17^2$. If $a, b, c, d$ are divisible by 7, there are positive integers $w, x, y, z$ such that
$$a = 7w, \quad b = 7x, \quad c = 7y, \quad d = 7z$$
and the given conditions on $a, b, c, d$ translate into
$$(i) w + x + y + z = 17^2$$
$$(ii) 17^2 \mid wx - yz$$
$$(iii) 17^2 \mid w^2 + x^2 + y^2 + z^2.$$
Multiplying (i) by $x$ and using (ii), we obtain
$$0 \equiv wx + x^2 + xy + xz \equiv yz + x^2 + yx + xz \equiv (x + y)(x + z) \pmod{17^2}.$$
Because $w > 0$, (i) implies that $0 < x + y < 17^2$ and $0 < x + z < 17^2$. Since 17 is a prime number, we now see that both, $x + y$ and $x + z$, must be divisible by 17. Knowing this, we can conclude from (i) that $w + z$ and

$w + y$ must also be divisible by 17. This means that $y \equiv z \equiv -w \pmod{17}$ and $x \equiv -y \pmod{17}$, hence $w^2 \equiv x^2 \equiv y^2 \equiv z^2 \pmod{17}$. Using (iii) we now see that
$$0 \equiv w^2 + x^2 + y^2 + z^2 \equiv 4w^2 \pmod{17}.$$
This implies that $w$ is divisible by 17, hence $w \equiv x \equiv y \equiv z \equiv 0 \pmod{17}$ as required.

Solution 2. As we are allowed to assume $a, b, c, d$ are divisible by 7, then we can write $a = 7a', b = 7b', c = 7c', d = 7d'$. Dropping the primes, and noting $2023 = 7 \cdot 17^2$ the problem requires us to investigate solutions to:
(i) $a + b + c + d = 17^2$
(ii) $17^2 \mid ab - cd$
(iii) $17^2 \mid a^2 + b^2 + c^2 + d^2$.
We are required to show that $a, b, c, d$ are all multiples of 17. It turns out the only relevant special properties of 17 are that it is an odd prime, so we make the following more general claim.
Let $p$ be an odd prime and let $a, b, c, d$ be positive integers such that
(i) $a + b + c + d = p^2$
(ii) $p^2 \mid ab - cd$
(iii) $p^2 \mid a^2 + b^2 + c^2 + d^2$.
Then all of $a, b, c, d$ are multiples of $p$. If we can prove this claim, the original problem is then solved by writing $p = 17$.
Adding or subtracting twice criterion (ii) from criterion (iii) gives
$$(a + b)^2 + (c - d)^2 \equiv 0 \pmod{p^2} \\ (a - b)^2 + (c + d)^2 \equiv 0 \pmod{p^2}.$$
Using (i) to replace $a+b$ in the first equation by $-c-d$, and $c+d$ by $-a-b$ in the second gives
$$c^2 + d^2 \equiv a^2 + b^2 \equiv 0 \pmod{p^2}.$$

Suppose for a contradiction that (without loss of generality given the symmetry of the equations) $a$ is not a multiple of $p$. Then, as $a$ and $p^2$ are relatively prime, there is a residue $i$ such that $b \equiv ai \pmod{p^2}$ and we have
$a^2 + b^2 \equiv a^2(1 + i^2) \equiv 0 \pmod{p^2}$ whence $i^2 = -1 \pmod{p^2}$. It is clear that $i \neq 0 \pmod{p}$ and $i \neq -1 \pmod{p}$ so that $ai \neq 0 \pmod{p}$ and $(1+i)a \neq 0 \pmod{p}$. Since $b \equiv ai \pmod{p^2}$ and $a+b \equiv (1+i)a \pmod{p^2}$, it follows that neither $b$ nor $a+b$ are multiples of $p$.
Now recall $c^2 + d^2 \equiv (d+ic)(d-ic) \equiv 0 \pmod{p^2}$. We cannot have both $p \mid d+ic$ and $p \mid d-ic$ as that would imply $p \mid d$ and hence $p \mid c+d$, contradicting $a+b+c+d=p^2$. So one of the factors must be divisible by $p^2$. By swapping $c$ and $d$ if necessary, we may assume $p^2 \mid d-ic$. Then we have $0 \equiv a+b+c+d \equiv (1+i)(a+c) \pmod{p^2}$.
As $i \neq -1 \pmod{p}$, we can deduce that $a+c \equiv 0 \pmod{p^2}$. Now $a, c$ are both positive integers so $a+c \ge p^2$. But then we have $0 < b+d = p^2-a-c \le 0$, which is the required contradiction.