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# Understand the problem

Let $p$ be a prime number. Find all positive integers $a,b,c\ge 1$ such that:
$$a^p+b^p=p^c.$$
##### Source of the problem
France IMO TST 2012, Problem 3
Number Theory
7/10
##### Suggested Book
Challenges and Thrills of Pre College Mathematics

Do you really need a hint? Try it first!

Observe that  we will try to fundamental solutions to $a^p + b^p = p^c$.  A fundamental solution (a,b,c,p) gives infinitely many solutions $(a.p^k, b.p^k, c+k, p)$. A fundamental solution is, therefore (a,b,c,p) if gcd(a,b) = 1.
We will focus on the fundamental solutions. We will deal with two cases: Case 1: p = 2 The equation reduces to $a^2 + b^2 = p^2$.  As gcd(a,b) = 1, it implies a and b are odd. Now any odd square = 1 mod 4. So, $a^2 + b^2 = 2 mod 4$. Hence, the only fundamental solution is (1,1,1) = (a,b,c)  We have that the following solutions are: $(1, 1, 1)$ and $\left(2^k, 2^k, 2^{2k+1}\right)$.

Case 2: p is an odd prime This now requires the idea of Lifting the Exponents. Please read here if you don’t know it. It is an advanced technique to deal with Diophantine Equations. Let’s check that the conditions of the LTE are satisfying here. p is an odd prime. gcd(a,b) = 1. p doesn’t divide a or b as we are looking for fundamental solutions. $a^p = a mod p; b^p = b mod p$. Hence, $a^p + b^p = a + b mod p$. So, p | a+b, and p don’t divide a or b.  Hence, we can apply LTE.

Using the LTE idea, we get  $1+v_p(a+b)=v_p\left(a^p+b^p\right) = c$ Then since $a+b | a^p+b^p$, we have that $a+b = p^{c-1}$, so $a^p+b^p = pa+pb$ Now, you see this can’t happen for large p, as the LHS is exploding too fast like exponential as p increases and RHS is linear in p. So, we will apply inequality to prove this and find a bound for p for which it works and search in that bound. Note that $a^p \geq pa, b^p \geq pb$ or $a^{p-1} \geq p, b^{p-1} \geq p$ if $a, b \geq 2$. Therefore, we must have that $a=1, b=p^{c-1}-1$ (or vice versa). But clearly $\left(p^{c-1}-1\right)^p > p^c$, so no solution. QED

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