<s>[INST] which of the following is true, number theory problem 
$a$ is correct as $n=2m+1 \text{(say)} \Rightarrow n^2=4m^2+4m+1\equiv1(\mod 8),\ m\ge1$
$b$ is true ingeneral, say $n=5, m=3$, for $c$ I dont know how to prove, please help
 [/INST] (a)$(2a+1)^2-1=4a^2+4a+1=8\frac{a(a+1)}2$
(b)$(2x+1)^2+(2y+1)^2=4(x^2+y^2+x+y)+2\equiv2\pmod 4$
But $(2z)^2\equiv 0\pmod 4,(2z\pm1)^2\equiv1\pmod 4$
(c)$$\frac{n^5}5+\frac{n^3}5+\frac{7n}{15}=\frac{n^5-n}5+\frac{n^3-n}3+\frac{7n}{15}+\frac n 5+\frac n 3$$
$$=\frac{n^5-n}5+\frac{n^3-n}3+n$$
Now using Femat's little theorem , prime $p\mid (n^p-n),$ so $5\mid (n^5-n)$ and 
$5\mid (n^3-3)$
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