For what values of $n \in \mathbb{N}$ does the two expressions $n^2 + 4n + 1 - p$ and $2n^2 + n + 2 - p$ ARE NOT perfect squares for any prime number $p$.

some progress - Manually I found out one such instance, where $n = 21$ there doesn't exist any prime number $p$ such that those two expressions could be perfect square.

We might solve each expression individually or both simultaneously.

EDIT : so peter pointed out that - For $n=21$, the second expression is a perfect square , for example , for $p=761$

Now adding the main background of this problem -

Conjecture 1 : For every $2n = a + b$, where $a, b, n \in \mathbb{N}$ there always exists minimum one pair $(a,b)$ such that $2(a+n) + ab$ is prime.

Conjecture 2 : For every $2n = a + b$, where $a, b, n \in \mathbb{N}$ there always exists minimum one pair $(a,b)$ with $gcd(a,b) = 1$ such that $n^2 + n + (ab + 2)$ is prime.

At this point I manually found the Exception to both the conjectures for $2n = 42$ implies $n = 21$. Even further supported by some numerical evidences.

Now lets substitute $b = 2n - a$ in the above expressions, with them resulting in some prime number say $p$

From first expression reduced to $a^2 -a(2n + 2) -2n + p$ we have its discriminant $\Delta = n^2 + 4n +1 -p$ which should must be perfect square.

Similarly for the second expression reduced to $a^2 -2an +p -n^2 -n -2$ we have its discriminant $\Delta = 2n^2 +n +2 -p$ which should must be a perfect square.

NOTE : Above both are quadratic expressions in $``a"$

so meanwhile translating the main problem to this very reduced format, I missed very essential constraint of Conjecture 2 , that is - $gcd(a,b) = 1$ and now when peter pointed out that it fails for $p = 761$ ,surprisingly there are other primes as well...

so lets check if our $gcd(a,b) = 1$ criteria is met or not, After plugging $n = 21$ and $p = 761$ in the expression $a^2 -2an +p -n^2 -n -2$ it boils down to solve for a quadratic equation in $``a"$ hereby - $a^2 -42a +297 = 0$ having roots as $a = 9, 33$

Finally $a = 9$ and $b = 33$ for $n = 21$ in the second expression, here clearly $gcd(a,b) = 3$

Hence my bad I couldn't induce this constraint into the problem you faced in the very beginning.

How should we approach this now, to conclude a rigorous solution.

  • 1
    $\begingroup$ For $n=21$, the second expression is a perfect square , for example , for $p=761$ $\endgroup$
    – Peter
    Feb 1, 2021 at 14:52
  • 1
    $\begingroup$ can we make a progress considering the actual missing constraint $gcd(a,b) = 1$ ?! $\endgroup$
    – Alphatrion
    Feb 1, 2021 at 16:10

1 Answer 1


For $\space n=21\space $, $2n^2 + n + 2 - 5=900=30^2.\space $ Also, $p=229\rightarrow 26^2\quad p=421\rightarrow 22^2\quad p=709\rightarrow 14^2\quad p=761\rightarrow12^2 \quad 941\rightarrow 6^2\quad $ There are also five "square" solutions to $n^2 + 4n + 1 - p,n=21$

The following is not proof. The following numbers work for $\space 2\le p \le 997\space $ but this simply indicates that the "square" solutions get rarer with altitude. For any of these, there may exist a larger $\space p\space$ for which either function will yield a perfect square.

$\space n\in\{59,71,75,96,98,101,120,121,122,124,126,129,133,134,136,138,140,145,\\146,149,156,164,170,171,173,174176,179,180,181,184,185,191,194,196,197,\\198,204\cdots\} \space$

I'm suggesting that there may be no value of $n$ meets your criteria.


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