1
$\begingroup$

If $k \equiv 1 \pmod 4$ and $q > 3$ (where $q$ is prime), does it follow that ${q^k}\sigma(q^k) \equiv q\sigma(q) \equiv 2 \pmod {q - 1}$?

Observe that $q\sigma(q) \mid {q^k}\sigma(q^k)$ when $k \equiv 1 \pmod 4$.

I got $q\sigma(q) \equiv 2 \pmod {q - 1}$ (for $q > 3$) from this Wolfram link.

Edit: After taking to anon, I would like to add that, for the problem I am considering, I actually have $q \equiv k \equiv 1 \pmod 4$. anon's answer gave $k \equiv 1 \pmod {q - 1}$ as a condition to check for the validity of the conjecture above.

1 Answers 1

4

Given $q$ is prime, we can use the fact that $q\equiv (q-1)+1\equiv 1\mod (q-1)$

in order to obtain the slightly more general result (for any nonnegative integers $r,k\ge0$)

$\begin{array}{c l} q^r\sigma_1(q^k) & =q^r\big(q^0+q^1+q^2+\cdots+q^k\big) \\[3pt] & \equiv (1)^r(\underbrace{1+1+1+\cdots+1}_{k+1}) \\ & \equiv k+1 \mod (q-1).\end{array}$

Of course plugging in $k=1$ and $r=1$ gives $q\,\sigma_1(q)\equiv 2\mod (q-1)$ as in the Wolfram link. The only reason the link says $q\ne3$ is because then the symbol '$2$' actually means $0$ modulo $3-1=2$.

(This means the conjecture is not generally true unless $k\equiv1\bmod (q-1)$ happens to hold.)

  • 0
    @anon, I think that divisibility constraint should be $\displaystyle\frac{q-1}{4}\mid\displaystyle\frac{k-1}{4}$. Anyway, yes - I agree that nothing further can be said about $\displaystyle\frac{q-1}{4}$ and $\displaystyle\frac{k-1}{4}$ without additional information about $q$ and $k$. Thanks for your help! :)2012-06-22