Problem 674

25. The side lengths of a triangle with integer sides and perimeter $n$

Let the side lengths be represented by $x, y, z$, then
$x+y+z=n$,
$x>0, y>0, 2x>y$.

The integer solutions for $x$ are given by $y=n-2x$, thus
$\frac{n}{4}<x<\frac{n}{2}$.

Therefore, the integer solutions for $x$ are $\left[\frac{n-1}{2}\right]-\left[\frac{n}{4}\right]$, i.e.,
$x=\left[\frac{n}{4}\right]+1,\left[\frac{n}{4}\right]+2, \cdots,\left[\frac{n-1}{2}\right]$.
Hence, $I=\left[\frac{n-1}{2}\right]-\left[\frac{n}{4}\right]$.
Now, let's find $T$, with $n=12q+r, 0 \leqslant r \leqslant 11$.
If $r=0$, then
$N=\frac{n^{2}-6n+8}{8}$,
$I=\frac{n}{2}-1-\frac{n}{4}=\frac{n-4}{4}$,
$E=1$,
$T=\frac{n^{2}-6n}{48}+8+\frac{n-4}{8}+\frac{1}{3}$
$=\frac{n^{2}}{48}$.
If $r=1$, then
$N=-\frac{n^{2}-1}{8}$,
$I=\frac{n-1}{2}-\frac{n-1}{4}=\frac{n-1}{4}$,
$E=0$,
$T=\frac{n^{2}-1}{48}+\frac{n-1}{8}$
$=\frac{n^{2}+6n-7}{48}$.
Similarly, we can obtain
If $r=2,3,4,5,6,7,8,9,10,11$,
then $T=\frac{n^{2}-4}{48}, \frac{n^{2}+6n+21}{48}$,
$\frac{n^{2}-16}{48}, \frac{n^{2}+6n-7}{48}$,
$\frac{n^{2}+12}{48}, \frac{n^{2}+6n+5}{48}$
$-\frac{n^{2}-16}{48}, \frac{n^{2}+6n+9}{48}$,
$\frac{n^{2}-4}{48}, \frac{n^{2}+6n+5}{48} \cdots$.
In summary, we have
When $n$ is odd, $T=N\left(\frac{n^{2}+6n}{48}\right)$;
When $n$ is even,
$T=N\left(\frac{n^{2}}{48}\right)$.

Where $N(x)$ represents the integer closest to $x$, for example.