# Quick problem?

I was just browsing for some small problems the other day and came across this problem and I am unsure if it should be obvious and have a quick answer. In any case, I couldn't figure it out.

If $$p$$ is a prime that divides $$a - b$$, then show that $$p^2$$ divides $$a^p - b^p$$, where $$a, b \in \mathbb{Z}$$.

Nevermind I figured it out using binomial theorem. But it looked like to me that $$p$$ doesn't necessarily have to be prime. Am I correct?

I have a solution. Let a = b + n*p

Then
$a^p - b^p = (b + np)^p - b^p = \sum_{k=1}^p \binom{p}{k} n^k p^k b^{p-k}$

Let's now consider each of the terms. For k = p, then it's pp, and since p >= 2, this is evenly divisible by p2. For the remaining terms, 1 <= k <= p-1, and from that inequality, the binomial coefficient is divisible by p, pk is also divisible by p, and thus, all these terms are divisible by p2.

Thus, all the terms are divisible by p2, implying that the whole expression is also.

Nevermind I figured it out using binomial theorem. But it looked like to me that $$p$$ doesn't necessarily have to be prime. Am I correct?
That's not the case. That is because binomial(n,k) for 1 <= k <= n-1 is always evenly divisible by n if n is a prime. For n composite, that need not be the case.

Counterexample: binomial(4,2) = 6, which is not divisible by 4.

In general for an odd prime: (a^p-b^P) =(a-b)(a^(p-1)+(a^p-2)b +a^(p-3)(b^2)+++(a^0)b^(p-1). Since there are p terms in the second expression and a==b Mod p was given, the second term goes to p(a^(p-1))==0 Mod p.

In fact, for the above situation, we need only an odd integer, not a prime.

Last edited:
robert, i like your expansion of a^p - b^p....but I'm afraid I don't understand why exactly it has to be an odd integer. Are you saying that there will not be p terms on the right hand side something if p is even? You started with p odd prime, does that mean the original question does not work for p = 2?

It's very easy to test. Let a = b + 2*k. I find

a2 - b2 = (b+2*k)2 - b2 = 4*k2 + 4*k*b = 4*k*(k+b)

4 = 22, of course.

ETA: robert Ihnot's argument will work for p being any integer >= 2.

ETA: robert Ihnot's argument will work for p being any integer >= 2.

Right that's what I thought, which is why I got confused when he said that we only need odd integers.

Forgive me if I am off here, but since we know that p divides a - b, then isn't this question essentially equivalent to stating, by Fermat's Little Theorem:

Can...

p (mod 2) divide a (mod p) - b (mod p)?

p divides n*mod (p) for any n
p divides a - b
and p (mod 2) , which equals (or is congruent to..) 1, clearly divides a (mod p) - b (mod p)

?

Last edited:
Forgive me if I am off here, but since we know that p divides a - b, then isn't this question essentially equivalent to stating, by Fermat's Little Theorem:

Can...

p (mod 2) divide a (mod p) - b (mod p)?

p divides n*mod (p) for any n
p divides a - b
and p (mod 2) , which equals 1, clearly divides a (mod p) - b (mod p)

?
I don't know exactly what you mean, but I think you're saying that if gcd(a,p)=gcd(b,p)=1, then by fermat

$$a^p\equiv\ a\ mod\ p\text{, and }b^p\equiv\ b\ mod\ p\ ==>\ a^p-b^p\equiv\ a-b\ mod\ p$$

it is always true only if p is prime or a carmichael number

it is always true only if p is prime or a carmichael number

I believe the assumption here is that p is prime. Honestly, however, I was not necessarily thinking in terms of gcd, but, rather, in the case of p being prime, then the above statements ought to cover all cases (i.e. permutations) of how a, b, and p can combine.

If I am not off-base here, one need only think about how p can "interact" with a - b, (a - b)(mod p) = n (mod p) & p(mod 2) can "interact" with a(mod p) - b(mod p).

- RF

P.S. Sorry, but the Latex editor does not work with my computer system.

Last edited:
you can use the identity $$a^p-b^p\equiv (a-b)^P\ mod\ p$$

of coursely if $$p|a-b\ ==>\ a^p-b^p\equiv\ 0\equiv\ (a-b)^P\ mod\ p$$

if $$p \geq 2,\ p|a-b ==> p^2|p^p|(a-b)^p$$

now you can conclude that $$p^2|a^p-b^p$$ with some observation

A trivial point, but...
if $$p \geq 2,\ p|a-b ==> p^2|p^p|(a-b)^p$$

p > or = to 2 is unnecessary to state since all p, by common definition, are > or = to 2.

Thank you, al mahed, for presenting in clear manner that which I was trying to get across in a less clear manner.

- RF

We know that (a-b)^p ==a^p-b^p Mod p. That is not the same as showing that (a-b)^p ==a^p-b^p Mod p^p, as you are suggesting above.

It is not the case that 10^7-3^7 is divisble by 7^7, in fact, 49 is the highest 7 divisor.

Last edited:
Hi Robert (see below post),

I realize I completely reversed terms here in rather dyslexic manner and so, in response to your post, and in the interests of not virally spreading misinformation, I am simply replacing what I posted previously with my initial surmise, which I believe was of value.

Forgive me if I am off here, but since we know that p divides a - b, then isn't this question essentially equivalent to stating, by Fermat's Little Theorem:

Can...

p (mod 2) divide a (mod p) - b (mod p)?

p divides n*mod (p) for any n
p divides a - b
and p (mod 2) , which equals (or is congruent to..) 1, clearly divides a (mod p) - b (mod p)

Last edited:
p^2 divides p^(a - b), which is congruent to p^a - p^b.

By Fermat's Litle Theorm, Modulus p. Fermat's Little Theorem says nothing about modulus P^2.

This is a weird coincidence since i was trying to solve a similar problem which is:

Prove that if a ^ p + b ^p = 0 (mod p) then a ^ p + b ^p = 0 (mod p ^ 2)

The logic is exactly similar i guess except that p as to be odd & greater than 2. Can somebody please look into & give me clues to solving the problem i posted a couple of days back with the subject line similar to "reduced residues" involving the Euler phi function? I would really appreciate it.

--
Sachin

Edit

Last edited:
Robert, what you want to prove is: $$a^p\equiv\ b^p\ mod\ p^2$$

Fermat don't say nothing about p^2 modulo, but euler do, where did you see modulus p^p in my post??

generally, if I'm not mistaken, $$a^{p^e}-b^{p^e}\equiv\ (a-b)^{p^e}\ mod\ p^e$$

you can use the identity $$a^p-b^p\equiv (a-b)^P\ mod\ p$$

of coursely if $$p|a-b\ ==>\ a^p-b^p\equiv\ 0\equiv\ (a-b)^P\ mod\ p$$

if $$p \geq 2,\ p|a-b ==> p^2|p^p|(a-b)^p$$

now you can conclude that $$p^2|a^p-b^p$$ with some observation

Right above!

Robert, what you want to prove is: $$a^p\equiv\ b^p\ mod\ p^2$$

Fermat don't say nothing about p^2 modulo, but euler do, where did you see modulus p^p in my post??

generally, if I'm not mistaken, $$a^{p^e}-b^{p^e}\equiv\ (a-b)^{p^e}\ mod\ p^e$$

I would be helpful to give a reference or show a proof.

Here is a case for mod 9. Now since phi of 9 = 6, we need only raise these numbers to the cube, but if you like, use 9 instead of 3.

$$2^3+5^3 \equiv133\equiv7 Mod 9.$$
But $$7^3\equiv 1 Mod 9.$$

my source is

http://planetmath.org/encyclopedia/FrobeniusAutomorphism4.html [Broken]

Last edited by a moderator:
The point of the situation with regards to p^2, is that, for example p=3, we have terms like 9!/(3!*6!) in the binominal expansion of (a+b)^9. In the mention term = 84, we have only 3, not 9, as a divisor.

Fermat's Little Theorm can be shown by induction: 1^p ==1, (x+1)^p ==x^p+1, but this is because all the intermediate terms in the binomial expansion contain p, a prime, which does not divide out. BUT THIS DOES NOT CARRY OVER TO P^2.

The point of the situation with regards to p^2, is that, for example p=3, we have terms like 9!/(3!*6!) in the binominal expansion of (a+b)^9. In the mention term = 84, we have only 3, not 9, as a divisor.

Fermat's Little Theorm can be shown by induction: 1^p ==1, (x+1)^p ==x^p+1, but this is because all the intermediate terms in the binomial expansion contain p, a prime, which does not divide out. BUT THIS DOES NOT CARRY OVER TO P^2.

yes, agreed, I know that, I didn't say the proof was complete, in fact I thought it would take only a few steps more, but that's not the case (I was trying to complete the proof here using what I've done, but gave me trouble, it is more complicated, I think)