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I have come across numbers of the form $$b=1+10a+50a^2+125a^3+125a^4$$ where $a$ is a positive integer.

Looking at the prime factors of $b$, I am conjecturing that all prime factors of $b$ are $\equiv 1 \pmod 5$. But I cannot prove this. Looking at $b$ in $\bmod 5$, we see that $b \equiv 1 \pmod 5$. If $p$ is a prime divisor of $b$, then I cannot necessarily conclude that $p \equiv 1 \pmod 5$, for example $91 \equiv 1 \pmod 5$, but $91 = 7 \times 13$. However, I think with the way that $b$ is defined, it can be shown that its prime divisors are actually congruent to $1 \pmod 5$.

Here are the first few numerical data for the choices of $a$:

$$a=1, b = 311, \{ 311 \}$$ $$a=2, b=3221, \{ 3221 \}$$ $$a=3, b=13981, \{ 11,31,41 \}$$

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    $\begingroup$ After a bit of testing, it really looks like this conjecture runs deeper than I anticipated... $\endgroup$ Commented Feb 8, 2023 at 5:55
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    $\begingroup$ Verified till $a=10000$ using python $\endgroup$
    – D S
    Commented Feb 8, 2023 at 7:11
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    $\begingroup$ Hint : $$b=\frac{(5a+1)^5-1}{25a}$$ $\endgroup$
    – Peter
    Commented Feb 8, 2023 at 7:51

2 Answers 2

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Peter beat me to it in the comments, but here is a general result. Let $\Phi_n(x)$ denote the $n^{th}$ cyclotomic polynomial. This is the unique sequence of monic integer polynomials satisfying

$$x^n - 1 = \prod_{d \mid n} \Phi_d(x)$$

which exist and are unique by Möbius inversion. Then:

Claim: If $a$ is an integer, then all the prime factors $p$ of $\Phi_n(a)$ which do not divide $n$ are congruent to $1 \bmod n$.

This result can be used, among other things, to prove that there are infinitely many primes congruent to $1 \bmod n$ in an elementary way without the full strength of Dirichlet's theorem.

Sketch. Let $p$ be a prime not dividing $n$ which divides $\Phi_n(a)$. The idea is to show that $\Phi_n(a) \equiv 1 \bmod p$ if and only if $a \bmod p$ has multiplicative order $n$ in the group of units $\mathbb{F}_p^{\times}$. Since by definition $\Phi_n(a) \mid a^n - 1$ we have that $a^n \equiv 1 \bmod p$, so $a$ has order dividing $n$. Letting $f(x) = x^n - 1$, we have that since $f'(x) = nx^{n-1}$ is nonzero $\bmod p$ by hypothesis, the roots of $f(x) \bmod p$ all have multiplicity $1$, and since by definition

$$x^n - 1 = \prod_{d \mid n} \Phi_n(x)$$

it follows that $\Phi_d(a) \not \equiv 0 \bmod p$ for all proper divisors $d$ of $n$ (or else $a$ would be a root of multiplicity greater than $1$), hence that $a^d \not \equiv 1 \bmod p$ for all proper divisors $d$ of $n$. So $a \bmod p$ has order exactly $n$ as desired.

Since $\mathbb{F}_p^{\times}$ has order $p-1$, by Lagrange's theorem $p-1$ must be divisible by $n$. $\Box$

This result strongly suggests that your polynomial $f(x) = 1 + 10x + 50x^2 + 125x^3 + 125x^4$ is related to $\Phi_5(x) = 1 + x + x^2 + x^3 + x^4$ in some way. We can relate them as follows:

$$\begin{eqnarray*} 5f(x) &=& 5 + 10(5x) + 10(5x)^2 + 5(5x)^3 + (5x)^4 \\ &=& \frac{(5x + 1)^5 - 1}{5x} \\ &=& \Phi_5(5x + 1). \end{eqnarray*}$$

So $f(x) = \frac{(5x + 1)^5 - 1}{25x}$ as Peter says in the comments. So if $p \neq 5$ is a prime dividing $f(a)$ for some integer $a$ then $p$ divides $\Phi_5(5a + 1)$ and then the claim above shows that $p \equiv 1 \bmod 5$. Moreover, since $f(a) \equiv 1 \bmod 5$, $p$ is never equal to $5$.

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  • $\begingroup$ shouldn't it be $\prod\limits_{d \mid n} \Phi_d(x)$ and not $\prod\limits_{d \mid n} \Phi_n(x)$? Otherwise, everything else makes sense. thanks $\endgroup$
    – Josh
    Commented Feb 8, 2023 at 16:03
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As Peter's comment states,

$$b = \frac{(5a+1)^5-1}{25a}$$

Thus, for any prime $p \mid b$, then also $p \mid (5a+1)^5 - 1$. Using the multiplicative order, with

$$m = \operatorname{ord}_p(5a+1)$$

then $m \mid 5$. However, $m \neq 1$ since $(5a+1)-1 = 5a$ and $\gcd(5a, b) = 1$, so $m = 5$. Both Fermat's little theorem and Euler's totient function give that $(5a+1)^{p-1}\equiv 1\pmod{p}$, which means that

$$5 \mid p - 1 \; \; \to \; \; p \equiv 1 \pmod{5}$$

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