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Fermat's little theorem says:

If $p$ is prime, and $a$ is an integer with $p \nmid a$, then $a^{p - 1 } \equiv 1 \pmod{p}$.

And this is a small part that I extracted from the book:

Let $n = 4k + r$ with $0 \leq r \leq 3$. Then by Fermat's little Theorem, we have $$b^n \equiv b^{4k + r} \equiv (b^4)^kb^r \equiv 1^kb^r \equiv b^r\pmod{5} \text{for any integer b}.$$

And I guess the author applied FLT for $b^4 \equiv 1 \pmod{5}$ But we don't know that $b$ divides 5 or not, how could the statement above be true? Any idea?

Thanks,
Chan

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    $\begingroup$ If $5|b$, then $b^n \equiv b^r \equiv 0 \bmod 5$ $\endgroup$ – user17762 Feb 26 '11 at 22:04
  • $\begingroup$ Fermat's little theorem is equivalent to the statement that a^p is congruent to a mod p with no hypotheses on a. $\endgroup$ – Qiaochu Yuan Feb 26 '11 at 22:04
  • $\begingroup$ If $b$ is a multiple of $5$ then both sides are zero. Otherwise the argument applies. $\endgroup$ – Yuval Filmus Feb 26 '11 at 22:05
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The exact statement of what you wrote is slightly incorrect. If $r=0$ and $b=5$ then it does not hold.

Most likely the condition $\gcd (b,5)=1$ is missing.

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  • $\begingroup$ Thanks. I strongly believe the statement was since the problem was: Prove that $5|(1^n + 2^n + 3^n + 4^n )$, and I guess the author implicitly thought $b$ was only in the set {1,2,3,4} $\endgroup$ – Chan Feb 26 '11 at 22:15

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