# perfect square of the form $n^2+an+b$

Given arbitrary integers $a,b$ , can we find in a systematic way an integer $n$ such that $$n^2+an+b$$ is a perfect square ?

• Some context would be nice. – user312437 Jul 30 '16 at 18:44
• Sometimes there is no such $n$. – André Nicolas Jul 30 '16 at 18:58
• maybe try writing the square as $(n+c)^2$. – supinf Jul 30 '16 at 18:59
• Case $a$ odd let $n=\left.(a-1)^2\right/4-b$ and $n^2+a n + b=\left(n+\frac{a-1}{2}\right)^2$ – Lozenges Jul 30 '16 at 19:54

If $n^2+an+b=m^2$, you can multiply everything by $4$ to get

$$4n^2+4an+4b=4m^2$$ and then complete the square to get $$(2n+a)^2-a^2+4b=4m^2$$

After rearranging and factoring, this is equivalent to $$(2n+a+2m)(2n+a-2m)=a^2-4b = \Delta$$

So any such $n$ will give you a factor of $\Delta$; conversely, if we write $\Delta=pq$ as the product of two integers, we can hope to recover $n$ by solving the system of linear equations $$2n+a+2m=p\\ 2n+a-2m=q$$ for $m$ and $n$, but it's not guaranteed that the $m,n$ you obtain in this way will be integers.

These equations simplify into $$n=\frac{p+q-2a}{4} \\ m=\frac{p-q}{4}$$

That is, for any given $a,b$, a procedure for finding all $n$ that work is as follows:

• Compute $\Delta=a^2-4b$.
• Write $\Delta$ as the product of two integers $p,q$ in all possible ways.
• For each such factorization, compute $n=\frac{p+q-2a}{4}$, $m=\frac{p-q}{4}$. If these are both integers, this is a possible value of $n$. If they are not, it isn't.

As the other answer mentions, it is definitely possible that you will get no solutions.

For example, say $a = 5$, $b=2$. Then $\Delta=5^2-4(2)=17$, and so there are two possibilities to try:

• Take $p=17,q=1$; then $n=\frac{17+1-10}{4}=2$, $m=\frac{17-1}{4}=4$.
• Take $p=-17,q=-1$; then $n=\frac{-17-1-10}{4}=-7$, $m=\frac{-17-(-1)}{4}=-4$.

These correspond to the fact that $2^2+5(2)+2=16$ and $(-7)^2+5(-7)+2=16$ are perfect squares.

• Thank you very much , That's exactly what I was looking for – Mohamed Perdu Jul 30 '16 at 19:47

I'm not sure if you want an answer (i.e., yes or no) or a systemic way to find $n$, but the short answer is no.

Whenever $a\equiv 0$ and $b\equiv 2$ mod $4$ we want a perfect square of the form $n^2+2\pmod 4.$ But a perfect square is always $0$ or $1$ mod $4$, so then $n^2+2$ is either $2$ or $3$ mod $4$. But then this cannot be a perfect square.