Find all the values: a+b=c+d=ab-cd=4n for n>1 This question was given to my cousin who is in grade 9 in Kazakhstan, hence the solution should not involve any matrices, sets etc. and I cannot think of a simpler solution. Any help would be greatly appreciated!
Find all the values: a+b=c+d=ab-cd=4n for n>1

The solution is basically the same as the one by Thomas.
WLOG, we can assume $a \le b$ and $c \le d$. So $a,c \le 2n$.
From the equations, we have $b = 4n - a, d = 4n - c$, and
$$
4n = a (4n-a) - c (4n-c) = (4n - a - c) (a - c).
$$
But $a+c$ and $a-c$ share the same parity, so they have to be both even:
$$
\begin{aligned}
a + c &= 2 p, \\
a - c &= 2 q,
\end{aligned}
$$
or
$$
\begin{aligned}
a &= p + q, \\
c &= p - q,
\end{aligned}
$$
and
$$
n = (2n - p) \, q.
$$
This means $q|n$, and $2n-p|n$. Since $p + |q| \le \max\{a, c\} \le 2n$, $2n-p\ge |q|$.  This means $q$ is positive, and $n \ge |q| q = q^2$.  So the general solution is
$$
\begin{aligned}
a = 2n - \left(\frac{n}{q} - q \right), \\
b = 2n + \left(\frac{n}{q} - q \right), \\
c = 2n - \left(\frac{n}{q} + q \right), \\
d = 2n + \left(\frac{n}{q} + q \right).
\end{aligned}
$$
with $1 \le q \le \sqrt{n}$.  The positions of $a$ and $b$ can be exchanged, so are those of $c$ and $d$.

For example, $n = 25$, we have, $q = 1$ or $q = 5$,
$$
\begin{aligned}
q = 1: \quad & a = 26, \; b = 74, \; c = 24, \; d = 76. \\
q = 5: \quad & a = 50, \; b = 50, \; c = 40, \; d = 60.
\end{aligned}
$$
