# Can $4\cdot n!-4n+1$ be a perfect square when $n>4$?

When $$n=4,4\cdot n!-4n+1=4\times24-4\times4+1=9^2.$$

I wonder if $$4\cdot n!-4n+1$$ can be a perfect square when $$n>4$$?

I searched $$n$$ from $$5$$ to $$10000$$ but no qualified number was found.

I know that:

(1) $$n$$ is even, since $$-4n+1 \equiv1 \mod 8$$;

(2) if $$p|4n-1$$ and $$p\leq n/2$$ then $$p^2|4n-1$$, here $$p$$ is prime;

(3) if $$p \not|4n-1$$ and $$p\leq n$$ then $$\left (\frac{-4n+1}{p}\right) = 1$$, here $$p$$ is prime, $$\left (\frac{a}{p}\right)$$ is Jacobi Symbol.

• This is equivalent to $n((n-1)!-1)=m(m+1)$, maybe this helps (or maybe not). – YiFan Oct 12 at 11:52
• I think it's still not known whether $n!+1$ can be a square for $n$ other than $4,5,7$. See oeis.org/A085692 – so there can be very hard problems not a million miles removed from yours. – Gerry Myerson Oct 12 at 12:12
• $(n-1)!-1$ doesn't have smaller divisor than $n$. – Takahiro Waki Oct 12 at 12:26
• @GerryMyerson As far as I know this has been solved in the meantime, but how is that related to this question ? – Peter Oct 12 at 12:46
• checked up to $10^9$: for(n=1, 10^9, a= 1-4*n; t= 1; forprime(p=2, n, if(kronecker(a,p)==-1, t= 0; break()); ); if(t, b= 4*n!+a; if(issquare(b), print(n" "a" "b))); ) – Dmitry Ezhov Oct 13 at 1:21

If $$\nu_p(4n-1)< \nu_p(n!)$$ for all primes $$p \mid (4n-1)$$, then we have- $$x^2 = 4n!-4n+1=(4n-1)(\frac{4n!}{4n-1}-1)$$ and for every $$p \mid (4n-1)$$, we also have $$p \mid \frac{4n!}{4n-1}$$. Thus- $$\gcd\bigg(4n-1,\frac{4n!}{4n-1}-1\bigg)=1 \implies (4n-1)=x_1^2$$ However, this is impossible as $$4n-1 \equiv 3 \pmod{4}$$ and $$3$$ is not a quadratic residue modulo $$4$$.

Next, we try to find all values of $$n$$ for which $$\nu_p(4n-1) \geqslant \nu_p(n!)$$ for some odd prime $$p \mid (4n-1)$$. Let $$\nu_p(4n-1)=t$$ and $$4n-1=kp^t$$ for positive integers $$k$$ and $$t$$. We have- $$4n-1=kp^t \implies n=\frac{kp^t+1}{4}\geqslant p\bigg\lfloor\frac{kp^t+1}{4p}\bigg\rfloor$$

Thus, we use the first the first term in Legendre's formula to establish- $$\nu_p(n!) \geqslant\bigg\lfloor\frac{kp^{t}+1}{4p}\bigg\rfloor$$

We must then have- $$t\geqslant\bigg\lfloor\frac{kp^{t}+1}{4p}\bigg\rfloor$$

Using some grunt work checking, it is quite easy to see that the only cases for $$t>1$$ are: $$(t,k,p)=(2,1,3),(2,1,5),(2,1,7),(2,1,11),(2,2,3),(2,2,5),(2,3,3),(3,1,3)$$ and we have $$4n-1=kp^t$$, which shows that- $$4n-1 \in \{9,25,49,121,18,50,27,27\} \implies 4n-1\in \{9,18,25,27,49,50,121\}$$

As $$4n-1 \equiv 3 \pmod{4}$$, the only possibility is $$4n-1=27$$ showing $$n=7$$, which fails.

For $$t=1$$, we get $$1 \geqslant \big\lfloor\frac{k+1}{4}\big\rfloor$$. Thus, we must have $$k<7$$. As $$4n-1$$ is odd, so is $$k$$. Thus, we have the only possibilities being- $$4n-1 \in \{p,3p,5p,7p\} \implies n \in \bigg\{\frac{p+1}{4},\frac{3p+1}{4},\frac{5p+1}{4},\frac{7p+1}{4}\bigg\}$$

for some prime $$p$$.

Now, if $$4n-1=qp$$ where $$q \in \{3,5,7\}$$, since $$t=1$$, we have $$q \neq p$$. Then, we will have $$\nu_q(n!) \leqslant \nu_q(4n-1)$$ since otherwise, the expression will have an odd power of $$q$$ which is prime in all cases. It can easily be seen that this means $$p<8$$ which gives us the work of checking- $$4n-1 \in \{9,15,21,25,35,49\}$$ out of which the only values $$3 \pmod{4}$$ are $$15$$ and $$31$$, which give the cases $$n=4,8$$. Only $$n=4$$ works.

Finally, we are left with the case $$4n-1=p$$ i.e. $$n=\frac{p+1}{4}$$ where $$p \equiv 3 \pmod{4}$$ is a prime. We have- $$x^2=4\bigg(\frac{p+1}{4}\bigg)!-p$$

Consider a prime $$q \leqslant \frac{p+1}{4}$$ (Not as defined earlier). We have- $$\bigg(\frac{p}{q}\bigg)\bigg(\frac{q}{p}\bigg)=(-1)^{\frac{p-1}{2} \cdot \frac{q-1}{2}}$$

Since $$x^2 \equiv -p \pmod{q}$$ must be a quadratic residue, we have- $$\bigg(\frac{-p}{q}\bigg) = 1 \implies \bigg(\frac{p}{q}\bigg)=(-1)^{\frac{q-1}{2}} \implies \bigg(\frac{q}{p}\bigg)=(-1)^{\frac{p-3}{2}\cdot\frac{q-1}{2}}=1$$ as $$p \equiv 3 \pmod{4}$$.

This means that all odd primes till $$\frac{p+1}{4}$$ are quadratic residues modulo $$p$$. One can check the cases below to show $$p > 7$$ which shows that- $$4\bigg(\frac{p+1}{4}\bigg)!\equiv 0 \pmod{8} \implies x^2 \equiv -p \equiv 1 \pmod{8} \implies p \equiv 7 \pmod{8}$$

This shows that $$2$$ is also a quadratic residue modulo $$p$$. Since all primes till $$\frac{p+1}{4}$$ are quadratic residues, all numbers till $$\frac{p+1}{4}$$ are quadratic residues too.

This means that we already have $$\frac{p-3}{4}$$ cases where a quadratic residue is followed by another quadratic residue. However, for $$p \equiv 3 \pmod{4}$$, this must occur exactly $$\frac{p-3}{4}$$ times. This means that for the following cases, we will never have consecutive quadratic residues.

This can be proved incorrect with the first value succeeding $$\frac{p+1}{4}$$ that is $$a \equiv 2 \pmod{6}$$ as $$a$$ and $$a+1$$ have all prime factors less than $$\frac{p+1}{4}$$.

Thus, we have the only solution being $$n=4$$ in the positive integers.

• First sentence, why is it that prime $p$ dividing $4n-1$ implies $p$ divides $4n!/(4n-1)$? – Gerry Myerson Oct 13 at 11:48
• As $\nu_p(4n-1)<\nu_p(n!)$ – Haran Oct 13 at 12:02