2

I've attempted to solve this problem. I can only conclude that if $N!+p(N)=x^y$, since $p(N)>N$, $x$ must be formed by primes greater than $N$ except for $p(N)$. I don't know how to go on. By calculation, the number concerned can even be a prime.

Falcon
  • 111
  • 1
    What is the source of this problem? Is there any reason to imagine that it has a sensible solution? After all, the question of finding the squares of the form $n!+1$ is still open. – lulu May 08 '23 at 23:14
  • 1
    Presumably for a prime power $x^y$ you need $x$ to be an integer and $y>1$ to also be an integer. Can you find a prime number which divides $N!+p_N$ only once? – Henry May 08 '23 at 23:14
  • 3
    See OEIS 121926 - apparently it has been proved it is not a square – Henry May 08 '23 at 23:21
  • No perfect power for $1\le n\le 10^4$ – Peter May 09 '23 at 11:23
  • @lulu, $(n! + 1)^2= (n!)^2 + 2*n! + 1$. Now, $(n!)^2$ itself can be easily calculated. So what am I missing about the difficulty of finding the square of $(n!+1)$? – user25406 May 09 '23 at 11:25
  • 1
    @user25406 You misunderstood my comment. Brocard's problem, still open, asks for solutions to $n!+1=m^2$. For example, $4!+1=5^2$, and there are a couple other small examples, but it is unknown if there are infinitely many solutions or not. – lulu May 09 '23 at 11:29

1 Answers1

1

For $y$ even, we can easily prove a stronger statement: $$p(N)\text{ is a quadratic non-residue modulo }N!$$

Proof: For $N <= 9$, check computationally and confirm the statement. Hence we can assume $N > 9$.

Denote $p(N)$ as $p_{N}$. Assume $p_{N}$ is a quadratic residue modulo $N!$. Then, $p_{N}$ must be a quadratic residue modulo every prime power that's a factor of $N!$, and by extension, every prime that's a factor of $N!$.

Observe that $N > 9$ implies that $4 \vert N!$, so we have $p_{N} \equiv 1 \pmod{4}$. Thus we can discard those $N$ where $p_{N}$ does not satisfy this condition. By the law of quadratic reciprocity, for some prime $q$ and $p_{N} = 4k + 1$ for some positive $k$:

$$\left(\frac{p_{N}}{q}\right)\left(\frac{q}{p_{N}}\right) = (-1)^{2k\frac{q-1}{2}} = 1$$ $$\left(\frac{p_{N}}{q}\right) = \left(\frac{q}{p_{N}}\right)$$

Thus, for $p_{N}$ to be a quadratic residue modulo some prime $q$, $q$ must be a quadratic residue modulo $p_{N}$. Since $p_{N}$ must be a quadratic residue modulo every prime smaller than $N$, each of these primes in turn must be a quadratic residue modulo $p_{N}$.

However, $N > 9$ gives $p_{N} > 23$, meaning that there must be at least one quadratic non-residue that's smaller than $\sqrt{p_{N}}$. By extension, there must also be a prime non-residue that's smaller than $\sqrt{p_{N}}$. Since $\sqrt{p_{N}} < N$ for $N > 9$, there must be a prime $q$ < $N$ such that $q$ is a quadratic non-residue modulo $p_{N}$. So we reach a contradiction and confirm the statement.

For $y$ odd, I think a different approach must be used, as it doesn't seem that something as elementary as the above argument will hold water, even for special cases (e.g. $y = 3$). You can perhaps look at the ideas presented in On Polynomial-Factorial Diophantine Equations by Daniel Berend and Jørgen E. Harmse, although their focus seems to be on reducible polynomials, which gives them more structure to work with.

My two cents is that the constant term in the equation being the $n$th prime number provides no additional structure over arbitrary equations of form $N! + p = x^y$ with $p$ prime and relatively small. The main argument used in the proof for $y$ even is the fact that $p(N)$ is prime and smaller than $N^2$. You could swap $p(N)$ for any prime number that's smaller than $N^2$ and the argument would hold. It seems most likely that any proof for the $y$ odd case (or special cases thereof) would exploit the primality and size of $p(N)$ rather than any hypothetical special properties of the number.

ynwarcs
  • 141