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Web search: $p$ is a prime, $a$ is any integer for which $a^p-a$ is a multiple of $p$.
My research: I found that if $p$ is smaller than a then the remainder of $a^p/p$ is not going to be a which is contradict to the theorem.
Examples: $$p=2,a=17$$ $$17/2=8+1/2$$ $$17^2/2=144+1/2$$ Contradiction shown.

Arctic Char
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Pck Tsp
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  • The theorem that $17^2 - 17$ will be a multiple of $2$. And it is. $17^2 - 17= 289-17 = 272$ which is divisible by $2$. That does mean that $17^2 \div 2$ and $17\div 2$ will have the same remainder. And they do; that remainder is $1$. But it doesn't mean the remainder will be $17$. It only means that t the remainder will be congruent to $17$. And it is. The remainder of $17^2 \div 2$ is $1$. And $1 \equiv 17\pmod 2$. – fleablood Apr 23 '21 at 14:43
  • $x\equiv a\pmod{!p}!\iff! p\mid x-a,$ is not the same as $,x \bmod p = a$. It's the same as $,x\bmod p = a\bmod p,,$ i.e. they're congruent $!\iff!$ they have equal remainders. See the linked dupe for more on the relationship between the congruence relation and mod (remainder) operation, and related topics. – Bill Dubuque Apr 23 '21 at 16:34

2 Answers2

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AFAIK, there is no such condition in Fermat's little Theorem. The example you considered is $p=2, a=17$. This works perfectly fine: $a^p-a=17^2-17=272$, which is a multiple of $p=2$. There is no contradiction here.

  • $17^2-17$ is the difference of two odds so is even - you don't need to calculate its value to see that [but if you do it's easier to note it $= 17(17-1) = 17(8\cdot\color{#c00}2)]\ \ $ – Bill Dubuque Apr 23 '21 at 08:10
  • So when 17^2/2~=1mod2 or 17 mod 2 – Pck Tsp Apr 23 '21 at 13:36
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Fermat's Little Theorem says that $a^p\equiv a \pmod p$. That means $a^p \div p$ and $a\div p$ will have the same remainder. It does NOT mean that the remainder will BE $a$. Just that $a^p$ and $a$ will have the same remainder.

And if $p < a$ then of course the remainder will not be $a$ because the remainder, $r$, by definition is less than $p$ so $r < p < a$. BUT it will be true that $a = r + k\cdot p$ for some (in this case positive) integer $k$. (or in other words that $p$ will divide $a - r$.)

Your examples are not contradiction. The remainder of $17\div 2$ is $1$. And the remainder of $17^2 \div 2$ is also $1$. And so $17^2 \equiv 1 \pmod 2$ and $17\equiv 1 \pmod 2$ and $17^2 \equiv 17 \pmod 2$ and $17^2 - 17$ is a multiple of $2$.

Everything behaves exactly as the theorem predicted.

......

A more illustrative example might be a larger prime. $p = 7$ and $a=11$ say.

$11= 1\times 7 + 4$ so the remainder is $4$.

And $11^7 = 19487171= 2783881\times 7 + 4$ also has a remainder of $4$.

And that's confirms the theorem and $4 +7 = 11$ and $11 \equiv 4 \pmod 7$ so

$11^7 \equiv 11 \pmod 7$ and

$11^7 -11= 19487171 - 11 = 19487160 = 2783880 \times 7$.

fleablood
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