This question was asked in my number theory quiz and I was unable to solve it.

Prove that there exists infinitely many primes of the form 5k-1.

Professor was kind enough to give a hint to consider $5(n!) ^{2} -1$ .

I proved that any prime dividing $5(n!) ^{2} -1$ must be greater than n but can't think of anything. Even I can't think along the lines along $x^{2} \equiv a$ (mod p) as 5 is there along with square of n! .

It's my humble request to you to shed some light on this question.

It's a first course on number theory and contains elementary number theory only.

  • $\begingroup$ See: en.wikipedia.org/wiki/Wilson%27s_theorem $\endgroup$ – vvg Oct 21 '20 at 16:16
  • $\begingroup$ Are you meant to have determined the primes for which $\left(\frac 5p\right)=1$? $\endgroup$ – lulu Oct 21 '20 at 16:18
  • $\begingroup$ It would help if you explained what techniques you have available. If, say, you are familiar with quadratic reciprocity, it is not difficult to finish the argument from the given hint. Can't say that I see a sensible way to avoid the use of something like that (which, of course, does not mean that there isn't one). $\endgroup$ – lulu Oct 21 '20 at 16:22
  • $\begingroup$ No he doesn't mean $\big(\frac{5}{p}\big)=1$ He just means $p\equiv 4\pmod{5}$ $\endgroup$ – Vlad Oct 21 '20 at 16:25
  • $\begingroup$ @Vlad Not following. The "usual" argument, based on the hint, would be to remark that $5$ must be a square mod any prime dividing $a_n=5(n!)^2-1$ and then to note that $a_n$ can't only be divisible by primes of the form $5k+1$. I assumed the intent here was to follow the hint. Should add: I'd say that using Dirichlet here is serious overkill. $\endgroup$ – lulu Oct 21 '20 at 16:27

Assume that there were only finitely many primes in the form $5k-1$, say $p_1, p_2\cdots\cdots p_r$ and $p_r$ is the largest. Consider the number $$N=5(p_r!)^2-1$$ Let $p$ be a prime divisor of $N$. Then we have $$5(p_r!)^2-1\equiv0\pmod p$$ $$(5p_r!)^2\equiv 5\pmod p$$

Therefore we see that $\left(\frac{5}p\right)=1$. In particular, $p\equiv \pm 1\pmod5$.

Now, if all prime divisors of $N$ were of the form $p\equiv1\pmod5$, then $N$ would be of the form $5k+1$, which is, however, not the case.

We conclude that at least on prime divisor of $N$ is of the form $5k-1$. Note that such $p$ cannot be any of the $p_i$ above. At this point we have arrived at a new prime of the form $5k-1$. Contradiction! BANG!


One of the best theoremes ever:

Dirichlet's theorem: Given any $a,b\in\mathbb{N}$ such that $\gcd(a,b)=1$, there are infinitely many primes in the arithmetic progression $an+b$ (and $bn+a$, obviously); as a reformulation, there are infinitely many primes that are $\equiv b\pmod{a}$ (and viceversa, again).

Remember that, it is good.

As an alternative solution to your problem, Wilson's theorem is pretty good for proving that for any prime $p$ there exist intifitely many primes of the form $pk-1$, but cannot even be compared to Dirichlet.

  • 1
    $\begingroup$ Please give a link or sketch of the proof using Wilson's theorem that your refer to above (note that the proof in Student1058's answer does not use Wilson's theorem). $\endgroup$ – Bill Dubuque Oct 21 '20 at 22:00

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