Understand the problem

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

Start with hints

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

Watch video

Connected Program at Cheenta

Math Olympiad Program

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

Geometry of AM GM Inequality

AM GM Inequality has a geometric interpretation. Watch the video discussion on it and try some hint problems to sharpen your skills.

Geometry of Cauchy Schwarz Inequality

Cauchy Schwarz Inequality is a powerful tool in Algebra. However it also has a geometric meaning. We provide video and problem sequence to explore that.

RMO 2019 Maharashtra and Goa Problem 2 Geometry

Understand the problemGiven a circle $latex \Gamma$, let $latex P$ be a point in its interior, and let $latex l$ be a line passing through $latex P$. Construct with proof using a ruler and compass, all circles which pass through $latex P$, are tangent to $latex...

RMO 2019 (Maharashtra Goa) Adding GCDs

Can you add GCDs? This problem from RMO 2019 (Maharashtra region) has a beautiful solution. We also give some bonus questions for you to try.

Number Theory, Ireland MO 2018, Problem 9

This problem in number theory is an elegant applications of the ideas of quadratic and cubic residues of a number. Try with our sequential hints.

Number Theory, France IMO TST 2012, Problem 3

This problem is an advanced number theory problem using the ideas of lifting the exponents. Try with our sequential hints.

Algebra, Austria MO 2016, Problem 4

This algebra problem is an elegant application of culminating the ideas of polynomials to give a simple proof of an inequality. Try with our sequential hints.

Number Theory, Cyprus IMO TST 2018, Problem 1

This problem is a beautiful and simple application of the ideas of inequality and bounds in number theory. Try with our sequential hints.

Number Theory, South Africa 2019, Problem 6

This problem in number theory is an elegant applciations of the modulo technique used in the diophantine equations. Try with our sequential hints

Number Theory, Korea Junior MO 2015, Problem 7

This problem in number theory is an elegant application of the ideas of the proof of infinitude of primes from Korea. Try with our sequential hints.