4
$\begingroup$

Find all integers $n$ for which the expression $$n^4+6n^3+11n^2+3n+31$$ is a perfect square.


My approach :

First I simplified the expression and equated it to some square number $$(n^2+3n+1)^2 - 3(n-10) = k^2$$ Now if we try to remove the term "$3(n-10)$" then the equation will be satisfied. And it is done only be equating $n = 10$ . I don't know how to find other solution. Please help me in solving further.

$\endgroup$
7
$\begingroup$

Strategy: try to find two perfect squares such that the expression $f(n)=n^4+6n^3+11n^2+3n+31$ is strictly between them for all but finitely many $n$.

$(n^2+3n+1)^2$ is important to compute, as you also pointed out. It is $n^4+6n^3+11n^2+6n+1$.

So our impression is that $f(n)$ is close to this, and that if $f(n)$ is a perfect square, it should be the square of $n^2+3n+1$.

We prove this for all but finitely many $n$. More precisely, we prove that for all but finitely many $n$, we have $(n^2+3n)^2< f(n)< (n^2+3n+2)^2$. First part: $(n^2+3n)^2= n^4+6n^3+9n^2< n^4+6n^3+11n^2+3n+31 \Leftrightarrow 0< 2n^2+3n+31$. This holds for all $n\in \mathbb{Z}$ (the discriminant is negative). Second part: $n^4+6n^3+11n^2+3n+31< (n^2+3n+2)^2 = n^4 + 6n^3 +13n^2 + 12n + 4 \Leftrightarrow$ $0< 2n^2+9n-27$.

This inequality fails iff $-6\leq n\leq 2$. So unfortunately, you have to check if $f(n)$ is a perfect square for these nine values. (Easy calculation.) Once you are done with that, if $n$ is not among these nine values, then $f(n)$ is strictly between $(n^2+3n)^2$ and $(n^2+3n+2)^2$, so it has to be $(n^2+3n+1)^2$. This leads to a unary equation that you have already solved.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.