
5. The side-lengths $a, b, c$ of a triangle $A B C$ are positive integers. Let

$$
T_{n}=(a+b+c)^{2 n}-(a-b+c)^{2 n}-(a+b-c)^{2 n}+(a-b-c)^{2 n}
$$

for any positive integer $n$. If $\frac{T_{2}}{2 T_{1}}=2023$ and $a>b>c$, determine all possible perimeters of the triangle $A B C$.


## Solution

Upon simplification we obtain $T_{1}=8 b c$ and $T_{2}=16 b c\left(3 a^{2}+b^{2}+c^{2}\right)$. Hence

$$
\frac{T_{2}}{2 T_{1}}=3 a^{2}+b^{2}+c^{2}
$$

Since $a>b>c$ we have $404<a^{2}<675$ which is equivalent to $21 \leq a \leq 25$. Since $2023 \equiv 3(\bmod 4), a$ can't be even for otherwise

$$
b^{2}+c^{2}=2023-3 a^{2} \equiv 3 \quad(\bmod 4)
$$

which is impossible because $b^{2}+c^{2} \equiv k(\bmod 4)$ where $k \leq 2$. If $a=21$ then

$$
b^{2}+c^{2}=2023-3.21^{2}=7.10^{2}
$$

implying $b \equiv 0(\bmod 7)$ and $c \equiv(\bmod 7)$. But then $b=7 b_{1}$ and $c=7 c_{1}$ for some positive integers $b_{1}$ and $c_{1}$ and we get

$$
7\left(b_{1}^{2}+c_{1}^{2}\right)=10^{2}
$$

which is absurd because 7 does not divide $10^{2}=100$. If $a=25$ then

$$
b^{2}+c^{2}=48
$$

implying that 3 divides $b^{2}+c^{2}$ which is true if and only if 3 divides both $b$ and $c$. Thus there exist positive integers $b_{2}$ and $c_{2}$ such that $b=3 b_{2}$ and $c=3 c_{2}$. But then we obtain

$$
3\left(b_{2}^{2}+c_{2}^{2}\right)=16
$$

which is absurd because 3 does not divide 16. When $a=23$,

$$
b^{2}+c^{2}=2023-1587=436=20^{2}+6^{2}
$$

Note that $a=23, b=20$ and $c=6$ are side-lengths of a non-degenerate triangle $A B C$ whose perimeter is 49 .
