Equations

Suppose there exists an integer $p$ for which $p^3 + 7p = 14(n^2+1)$ .

$p^3+0 \equiv 0 \pmod{7}$ , so $7|p$ , so let $p=7k$ , where $k \in \mathbb{N}$ . So now

$343k^{3} + 49k = 14(n^2+1)$ , or

$7(7k^{3} +k) = 2(n^2+1)$ , so that

$n^{2} \equiv -1 \pmod{7}$ , which is not possible because $-1$ is not a quadratic residue modulo any prime of the form $4l+3$ .

Therefore $\alpha = 0$ , and $101\alpha = \boxed{0}$ .

We will show that the equation has no integral solution for any $n$ . Suppose $a$ is an integral root of the equation,then

$\displaystyle \begin{aligned} &a^3+7a-14(n^2+1)=0 \\&a^3=7(2n^2+2-a) \end{aligned}$

Hence $7\mid a$ ,so $a=7k$ for some integer $k$ . Putting back and simplifying we have $7k(1+49k^2)=2(n^2+1)$ . This suggests $7\mid (n^2+1)$ . $n$ can not be a multiple of $7$ clearly so $n$ can be of the forms $\{7m\pm 1,7m\pm 2 , 7m\pm 3\}$ . Simple arithmetic shows that : $(7m\pm 1)^2 +1\equiv 2\mod{7}$ , similarly $(7m\pm 2)^2 +1\equiv 5\mod{7}$ , $(7m\pm 3)^2 +1\equiv 3\mod{7}$ . So clearly $7\not\mid (n^2+1)$ for any $n$ which is a contradiction and thus there are no integral roots.

It might be interesting to note that $7\mid a^2+b^2 \implies 7\mid a \text{ and } 7\mid b$