Let $p$ be a prime number such that $p \equiv 1 \pmod 4$. Prove that $\frac{p^p-1}{p-1}$ is not prime.
We can rewrite $\frac{p^p-1}{p-1}$ as $$\dfrac{p^p-1}{p-1} = 1+p+p^2+\cdots+p^{p-1},$$but how do we show this is not prime?
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user19405892
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2Would be much easier if it were $\frac{p^{p-1}-1}{p-1}$. Or $\frac{p^{p+k}-1}{p-1}$ for any odd $k$, really. – Gabriel Burns Nov 03 '16 at 14:02
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2What have you tried? I factored the numbers with $p=5,13,17$ and didn't find any factors that were obviously related to $p$ – Ross Millikan Nov 03 '16 at 14:05
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1Possibly related: https://math.stackexchange.com/questions/223847/prove-that-all-divisors-of-fracpp-1p-1-are-of-the-form-pk1-where-p – Watson Nov 03 '16 at 14:23
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1@RossMillikan if $f(p) = \frac{p^p - 1}{p - 1}$, then $5\times 2 + 1 \mid f(5)$ and $13 \times 4 + 1 \mid f(13)$ and $17 \times 644 + 1 \mid f(17)$ ... the key seems to be to look at the problem in terms of $\pmod {p + 1}$, and it might help to keep in mind the the sum $\sum_{k = 1}^p p^k \equiv \sum_{k = 1}^p k \pmod{p + 1}$ because modular exponents form a bijection. – DanielV Nov 03 '16 at 14:26
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1Also $29 \times 2 + 1 \mid f(29)$ – DanielV Nov 03 '16 at 14:29
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@DanielV $(p^p-1)/(p-1)\equiv 1\pmod p$, so it must have a prime divisor of form $kp+1$, if it's not prime(in fact all divisor of $(p^p-1)/(p-1)$ is of form $kp+1$. – Cave Johnson Nov 03 '16 at 14:32
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it is also for first 20 primes :( – kotomord Nov 03 '16 at 14:34
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Worth noting that the sum is $\equiv 1 \mod p$ and $\mod p+1$ and $\mod p^3+1$ – Gabriel Burns Nov 03 '16 at 14:38
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and $\mod p^2+1$ – Gabriel Burns Nov 03 '16 at 14:52
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and thus also mod any product of those less than itself. – Gabriel Burns Nov 03 '16 at 14:57
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@GabrielBurns : why would it be easier for $(p^{p-1}-1)/(p-1)$ ? – Alphonse Nov 03 '16 at 16:47
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1@Alphonse because that would be a summation with $p-1$ terms, all congruent to $1 \mod 4$. Since $p-1 \equiv 0 \mod 4$, the result would be a multiple of $4$. – Gabriel Burns Nov 03 '16 at 16:53
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This might be useful: https://en.wikipedia.org/wiki/Fermat%27s_theorem_on_sums_of_two_squares – Gabriel Burns Nov 03 '16 at 17:04
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3@user Where do you get this stuff? – Will Jagy Nov 03 '16 at 20:33
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We may notice that $\frac{p^p-1}{p-1}=\Phi_p(p)$, where $\Phi_p$ is the $p$-th cyclotomic polynomial. If we assume that for some prime $q$ we have $\Phi_p(x)\equiv 0\pmod{q}$, then $x$ has order $p$ in $\mathbb{Z}/(q\mathbb{Z})^*$, hence $p\mid(q-1)$, or $q\equiv 1\pmod{p}$, by Lagrange's theorem. Additionally, the constraint $p\equiv 1\pmod{4}$ ensures that $\Phi_p(p)$ has a Aurifeuillean factorization.
Jack D'Aurizio
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@user19405892: I guess so, since a Aurifeuillean factorization ensures that $\Phi_p(p)$ cannot be prime. – Jack D'Aurizio Nov 03 '16 at 18:53
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For short, the fact that $p$ splits in $\mathbb{Z}[i]$ implies that $\Phi_p(p)$ splits over $\mathbb{Z}$. – Jack D'Aurizio Nov 03 '16 at 18:54
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1Where can I find a reference about proofs of which cyclotomic polynomials have an Aurifeuillean factorization ? – Oct 17 '17 at 15:14
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In the linked Wiki page only the fact is stated and the documents in the reference, further reading section doesn't seem to contain any proof – Oct 17 '17 at 17:37
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@misao: Aurifeullean factorizations are ultimately a consequence of Gauss sums: https://en.wikipedia.org/wiki/Gauss_sum – Jack D'Aurizio Oct 17 '17 at 17:40
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1@misao I found Paul Garrett's notes on Aurifeuillean stuff useful. Note, Paul Garrett also contributes here occasionally. – Jyrki Lahtonen Oct 21 '17 at 05:41