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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 ?

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

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

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