Hey I recently got interested in number theory and proved the following inequality:

\begin{align} x^m-y^m > p_n^2 + 4p_n + 3 > 1\ \text{(corrected again)} \end{align}

where $x-y\neq1$ and m in an integer >1


\begin{align} \gcd(x,y)&=1 \\ xy&= 2*3*5*..*p_n \end{align}

So my question is ... Does this formula already exist? And is it useful or slightly interesting?

  • $\begingroup$ Presumably you also assume $x>y$? $\endgroup$ Sep 16, 2013 at 12:20
  • $\begingroup$ I thought that was self implied as x^m>y^m+1 ... so x^m>y^m hence, x>y $\endgroup$
    – drewdles
    Sep 16, 2013 at 12:31
  • 1
    $\begingroup$ No, this is not self implied! You write $x-y \ne 1$. $\endgroup$ Sep 16, 2013 at 12:39
  • $\begingroup$ well I say g.c.d (x,y) =1 ... Hence x and y must be integers $\endgroup$
    – drewdles
    Sep 16, 2013 at 12:43
  • 2
    $\begingroup$ @Jyrki: Your $A(n)$ is IMO the primorial function $\#p_n$ and the asymptotic growth rate is $\#p_n \approx e^{\#p_n}$, see mathworld.wolfram.com/Primorial.html formula (4). $\endgroup$ Sep 16, 2013 at 12:51

2 Answers 2


The inequality deals with the primorial function $$ p_n\#=2\cdot3\cdot5\cdots p_n=\prod_{p\le p_n,\ p\in \mathbb{P}}p, $$ where $p_n$ is the $n$th prime. Asymptotically we have the result that $$ \lim_{n\to\infty}\frac{\ln p_n\#}{p_n}=1. $$ Early on the primorials are bit smaller though. For example $\ln(59\#)\approx49$.

Consider the following problem. Assume that $xy=p_n\#$ and that $x-y\ge3$ (in OP $x,y$ were constrained to be integers of opposite parity such that $x-y>1$ implying that $x-y\ge 3$). Therefore $$x^m-y^m\ge x^2-y^2=(x-y)(x+y)\ge3(x+y).$$ Here by the AM-GM inequality $x+y\ge2\sqrt{xy}=2\sqrt{p_n\#}.$ Therefore asymptotically we get a lower bound $$ x^m-y^m\ge 6\sqrt{p_n\#}\ge6e^{\frac n2(1+o(1))}. $$

Asymptotically we also have have $p_n\approx n\ln n.$ This suggests that $$ \frac{\ln(x^m-y^m)}{\ln p_n}\ge \frac n{2\ln n} K(n), $$ where $K(n)$ is some correction factor (bounded away from zero) that I won't calculate.

Your result says (using only the main term $p_n^2$) that $$ \frac{\ln(x^m-y^m)}{\ln p_n}\ge 2. $$ So asymptotically it is weaker. But it would not be fair to call your result trivial because of this. I'm not a number theorist, but I have seen simpler estimates being derived in many number theory books, and in addition to being fun, they pave the road to stronger results.

Please share details of your argument with us, so that we can comment and give you other kind of feedback!

  • $\begingroup$ Asymptotically it is, of course, irrelevant whether I require $x-y\ge1$ or $x-y\ge3$ here. $\endgroup$ Sep 16, 2013 at 13:36
  • $\begingroup$ this is gonna b a long story $\endgroup$
    – drewdles
    Sep 16, 2013 at 13:41
  • $\begingroup$ @Anant: I think you can answer your own question also. Then you won't be constrained by the 500 character limit. $\endgroup$ Sep 16, 2013 at 13:49
  • $\begingroup$ Sorry about the bad mistakes in the definitions. They were a result of earlier editing, where I tried to avoid using subscripts on primes. $\endgroup$ Sep 20, 2013 at 4:16
  • $\begingroup$ Does $ \ln p_n \# < p_n $ always hold? If so, where can a proof be found? $\endgroup$ Apr 9, 2015 at 5:44

Modulo small values of $n,$ you do not need to assume that $x-y\ne 1$ as well as asymptotic on the primorial function. Instead, you can get away with Bertrand's postulate stating that $p_{n}<2p_{n-1}$ or $p_{n-1}\ge p_n/2.$ Indeed, if $m\ge 2,$ then $$x^m-y^m=(x-y)(x^{m-1}+...+y^{m-1})\ge (x-y)(x+y)\ge x+y\ge 2\sqrt{xy}.$$ Note, $$2\sqrt{xy}=2\sqrt{p_1...p_{n-3}p_{n-2}p_{n-1}p_n}$$ and using the fact that $p_{n-2}\ge p_n/4,$ $p_{n-3}\ge p_n/8$ and $p_{n-1}\ge p_n/2$ we can estimate $2\sqrt{xy}\ge 2\sqrt{p_1p_2\cdot...\frac{p_n^4}{64}}=\frac{p_n^2}{8}\sqrt{p_1p_2...}.$ So we are left to check the result for small values of $p_n.$


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