Squares

Rewrite the equation as: $t^4+4t^3+t^2+2t+25 =a^2$

Let $f(t) = t^4+4t^3+t^2+2t+25$ : our goal consists in finding $t$ for which $f(t)$ is a square. We can write:

$\bigg \{ \begin{array}{rl} f(t) > (t^2+2t-3)^2 & t \ne -2 \hspace{5.04cm} (1) \\ f(t) < (t^2+2t+1)^2& t \ne -2,-1,0,1,2 \hspace{3cm} (2) \\ \end{array}$ $\rightarrow (t^2+2t-3)^2 < f(t) < (t^2+2t+1)^2 \hspace{3cm} t \ne -2,-1,0,1,2$

The value for which $t$ does not satisfy the inequality can be find solving each inequality. For instance the $(2)$ : $t^4+4t^3+t^2+2t+25 > (t^2+2t+1)^2$ $\rightarrow 5t^2+2t-24>0$ $\rightarrow t< -\frac{12}{5} \vee t>2$ implying that $t \ne -2,-1,0,1,2$ . From the inequality $(t^2+2t-3)^2 < f(t) < (t^2+2t+1)^2$ we know that for almost all values of $t$ , $f(t)$ is a number between two square which differ from 4. Since $f(t)$ must be a square, we have only 3 possibilities:

• $f(t) = (t^2+2t-2)^2$ that has solutions $t=-3 \vee t=-7$

• $f(t) = (t^2+2t-1)^2$ that has no solutions in $\mathbb{Z}$

• $f(t) = (t^2+2t)^2$ that has no solutions in $\mathbb{Z}$

Finally, we have to check by hand the cases $t = -2,-1,0,1,2$ .

The solutions are $(a,t) = (\pm 1, -39), (\pm 3, -2), (\pm 5, 0), (\pm 9, 2), (\pm 33,-7)$ , so the answer is $\boxed{10}$

There are 10 solutions.