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$a$ and $b$ are natural numbers and $p$ is a prime number. Prove that if $a^{p}-b^{p}$ is divisible by $p$, then it is also divisible by $p^{2}$.

My attempt: Based on Fermat's theorem $(a^{p}-a)$ and $(b^{p}-b)$ are divisible by p, therefore their difference is divisible by $p$. I.e. $(a^{p}-a)-(b^{p}-b)=(a^{p}-b^{p})-(a-b)$ is divisible by $p$. Since the left hand side of this equation is divisible by $p$, the right hand side also should be divisible by $p$. Based on the problem's assumption $(a^{p}-b^{p})$ is divisible by $p$ therefore we conclude $(a-b)$ must be divisible by $p$. Now We can factorize $(a^{p}-b^{p})$ $$(a^{p}-b^{p})=(a-b)(a^{p-1}+a^{p-2}b+....+b^{p-1})$$ we have to prove $(a^{p}-b^{p})$ is multiple of $p^2$ which means we have to show the right hand side should be multiple of $p^{2}$. But $(a-b)$ is multiple of p therefore we have to show $(a^{p-1}+a^{p-2}b+\cdots+b^{p-1})$ is multiple of $p$ and I am stuck here. Any help would be appreciated.

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    $\begingroup$ "$x$ divides $y$" means that there is some $z$ such that $xz=y$. Are you sure that this what you mean? Or perhaps the opposite? $\endgroup$ – vadim123 Mar 16 '15 at 15:14
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    $\begingroup$ This problem shows it is true for odd primes. math.stackexchange.com/questions/783274/if-p-mpnp-prove-p2-mpnp/… $\endgroup$ – Thomas Andrews Mar 16 '15 at 15:23
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    $\begingroup$ You seem a bit mixed up about "divides" and "is divisible by". You have done all the hard work of (almost) proving that if $a^{p}-b^{p}$ is divisible by the prime $p$, then $a^{p}-b^{p}$ is divisible by $p^{2}.$ Note that in the rightmost factor of your expression, there are $p$ terms..... $\endgroup$ – Geoff Robinson Mar 16 '15 at 15:26
  • $\begingroup$ @vadim123 a{p}-b{p}is multiple of p{2} $\endgroup$ – max Mar 16 '15 at 15:35
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To start with your beginning.

Knowing that $p\mid a-b$ means that $a\equiv b\pmod p$. So $a^kb^{p-1-k}\equiv a^{p-1}\pmod p$. So:

$$a^{p-1}+a^{p-2}b+\cdots + ab^{p-2}+b^{p-1}\equiv a^{p-1}+a^{p-1}\cdots a^{p-1}=pa^{p-1}\equiv 0\pmod p$$

So both $a-b$ and $a^{p-1}+a^{p-2}b+\cdots + ab^{p-2}+b^{p-1}\equiv pa^{p-1}$ are divisible by $p$, and hence $a^p-b^p$ is divisible by $p^2$.

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I'm going to assume $p$ is an odd prime. $a^p\equiv a $ and $b^p\equiv b\bmod p$ therefore $(a-b)$ is a multiple of $p$. If $a$ is a multiple of $p$ then $b$ must also be and it becomes trivial to see both $a^p$ and $b^p$ are multiples of $p^2$. The other case is when neither number is a multiple of $p$, we then apply the Lifting The Exponent Lemma to conclude.

Notice if $p$ is $2$ then it is also true, because if $a$ is odd so is $b$ and odd squares are $1\bmod 4$. If $a$ is even so is $b$ and even squares are $0\bmod 4$.

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  • $\begingroup$ At the very least, you should state the lifting lemma in your answer, or else the answer is not self-contained. External links get stale. $\endgroup$ – Thomas Andrews Mar 16 '15 at 16:36
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    $\begingroup$ (But it looks like the lifting lemma is really overkill, and really assumes what is meant to be proven.) $\endgroup$ – Thomas Andrews Mar 16 '15 at 16:38
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Hint: Once you know that $p$ divides $a-b$, then $a=b+pk$ for some integer $k$. Try the binomial expansion on

$$a^p-b^p=(b+pk)^p-b^p$$

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