If p divides a - b, show p² divides a^p - b^p

[al-mahed]

Ah, I missed that, sorry.

don't worry

anyway, I was thinking on a similar problem, to prove (or disprove) the statement:p^5 never divides a^p+b^p, that is, the contrary proplem (a superior limit), ohh, and gcd(a,p)=(b,p)=1

 

[robert Ihnot]

Another way to approach this whole thing depends upon using the Taylor Theorem, but in a very truncated form. I will give an example:

F(x) = 2^5 + x^5 == 0. Mod 5. In this case, x=3. It is very easy to move to the case Modulus=25, which has already been show, first by lpetnich, and then by me: Using the Taylor method we arrive at:

F(x+hp) = F(x) +hpF'(x) +(hp)^2(F''(x)) ++++ But in this case we don't even need the term (hp)^2(F''(x)).

BUT since F'(x) =5x^4, we already have arrive at the p^2 = 25 situation, and we are left with just F(3) = 2^5+3^5 ==0 Mod 25.

However to give this proof some legitimacy, let's look at raising the example to Modulus 5^3 = 125, and using only the form 5kF'(x).
$$F(x+hp) = F(x) + hpF'(x) \equiv 0 Mod 125.$$
$$hpF'(3) \equiv -(2^5+3^5) Mod 125$$
Since F'(3) =5*3^4, and dividing out 25 from the numerator, we have: $$h=\frac{-2^5-3^5}{25*81} = -11/81=-1 Mod 5,$$.

Thus h=-1 and 3+(-1)*5 = =-2==123 Mod 125.

Thus moving up in p as an exponent, we have: $$2^{25}+123^{25} \equiv 0 Mod 125.$$

But since again we get the simplist case as before for Mod 25, the above case works for
$$2^{25}+123^{25}\equiv 0 Mod 625.$$

While it might look pulling a rabbit out of the hat, if we always raise by F(x)+h(p^k)F'(x) Modulus p^(k+1), which would have been the normal expectation for a higher power,we get always the simplist case; so that nothing is need but to raise the exponent:2+3==0 Mod5. 2^5+3^5 == 0 Mod 25, 2^25+3^25 ==0 Mod 125, 2^125+3^125 ==0 Mod 625, 2^625+3^625 ==0 Mod 3125.

Remember all these cases involve division by powers of p, they do not handle the case (a-b)^(p^k)==a^(p^k)-b^(p^k) Mod p^(k), as the first part of the Freshman's Dream indicates. The last part of the Freshman's Dream is nothing but a generalization of Fermat's Little Theorem. I thought I handled that by indicating proof by induction 1^p==1 Mod p, (k+1)^p ==k^p + 1 Mod p, because all middle terms in the binominal expansion contain p. Thus x^p==x Mod p for the multiplicative group mod p.

 

[robert Ihnot]

Check me out page 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.

Thanks in advance,
--
Sachin