With Bertrand's postulate at hand, it is easy to see that $n!$ is never a square for $n\ge 2$ (because there is a prime between $n/2$ and $n$).

But are there more elementary proofs of that fact?

  • $\begingroup$ See whether this helps you: quora.com/… $\endgroup$ – Rohan Dec 14 '17 at 14:25
  • $\begingroup$ Well, I'd hardly say that that proof is more elementary. $\endgroup$ – ajotatxe Dec 14 '17 at 14:30
  • $\begingroup$ Yes, you are right. $\endgroup$ – Rohan Dec 14 '17 at 14:30
  • 2
  • $\begingroup$ @mathlove This is near from what I'm looking for. But if you compaqre Erdős' proof of the Bertrands' postulate with this solution, you can note that they are almost identical. That is, this proof has an "embedded" proof the Bertrand's postulate. $\endgroup$ – ajotatxe Dec 15 '17 at 15:02

Not an answer just some speculative thoughts.

Let's define $T_n$ as the n'th tetrahedral number, where


The even tetrahedral numbers $T_{2n}$ are also given by


The odd tetrahedral numbers $T_{2n-1}$ are also given by


We can define the factorial in terms of a tetrahedral number thus

$$n!=6(n-3)! \;T_{n-2}$$

We can then iterate this process as long as the residual factorial left over is 2! or greater. The factorial is then calculated from the product of two or more tetrahedral numbers, alternating between odd and even tetrahedral numbers.

$$n!=6^2(n-6)! \;T_{n-5}\;T_{n-2}$$ $$n!=6^3(n-9)! \;T_{n-8}\;T_{n-5}\;T_{n-2}$$

According to Dickson (History of the Theory of Numbers (Vol II. p.25)) A.J.J. Meyl [REF1] has proved that only three tetrahedral numbers are perfect squares; that is $T_1=1^2$, $T_2=2^2$ and $T_{48}=140^2$. Assuming it is correct it is an elementary proof.

The question is whether it is possible to build on this proof and the properties of tetrahedral numbers, to determine when the product of odd and even tetrahedral numbers are squares or not squares.

[REF1] Meyl, A.J.J. "Solution de Question 1194." Nouv. Ann. Math. 17, 464-467, 1878. http://www.numdam.org/article/NAM_1878_2_17__464_1.pdf


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.