 # 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

# Connected Program at Cheenta

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

# Similar Problems

## Sum of Sides of Triangle | PRMO-2018 | Problem No-17

Try this beautiful Problem on Geometry from PRMO -2018.You may use sequential hints to solve the problem.

## Recursion Problem | AMC 10A, 2019| Problem No 15

Try this beautiful Problem on Algebra from AMC 10A, 2019. Problem-15, You may use sequential hints to solve the problem.

## Roots of Polynomial | AMC 10A, 2019| Problem No 24

Try this beautiful Problem on Algebra from AMC 10A, 2019. Problem-24, You may use sequential hints to solve the problem.

## Set of Fractions | AMC 10A, 2015| Problem No 15

Try this beautiful Problem on Algebra from AMC 10A, 2015. Problem-15. You may use sequential hints to solve the problem.

## Indian Olympiad Qualifier in Mathematics – IOQM

Due to COVID 19 Pandemic, the Maths Olympiad stages in India has changed. Here is the announcement published by HBCSE: Important Announcement [Updated:14-Sept-2020]The national Olympiad programme in mathematics culminating in the International Mathematical Olympiad...

## Positive Integers and Quadrilateral | AMC 10A 2015 | Sum 24

Try this beautiful Problem on Rectangle and triangle from AMC 10A, 2015. Problem-24. You may use sequential hints to solve the problem.

## Rectangular Piece of Paper | AMC 10A, 2014| Problem No 22

Try this beautiful Problem on Rectangle and triangle from AMC 10A, 2014. Problem-23. You may use sequential hints to solve the problem.

## Probability in Marbles | AMC 10A, 2010| Problem No 23

Try this beautiful Problem on Probability from AMC 10A, 2010. Problem-23. You may use sequential hints to solve the problem.

## Points on a circle | AMC 10A, 2010| Problem No 22

Try this beautiful Problem on Number theory based on Triangle and Circle from AMC 10A, 2010. Problem-22. You may use sequential hints to solve the problem.

## Circle and Equilateral Triangle | AMC 10A, 2017| Problem No 22

Try this beautiful Problem on Triangle and Circle from AMC 10A, 2017. Problem-22. You may use sequential hints to solve the problem.