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# Bangladesh MO 2019 Problem 1 – Number Theory

A basic and beautiful application of Numebr Theory and Modular Arithmetic to the Bangladesh MO 2019 Problem 1.

# Understand the problem

Find all prime numbers such that the square of the prime number can be written as the sum of cubes of two positive integers.
Number Theory
5/10
##### Suggested Book
A Friendly Introduction to Number Theory by J.H.Silverman

Do you really need a hint? Try it first!

Write the problem in a Mathematical Equation form i.e. $p^2 = a^3 + b^3$. Now can you like factorize the stuff to make life easier and use divisibility rules?
After factorizing, we get $p^2 = (a+b)(a^2 + b^2 - ab)$. Now can use the prime factorization idea and see what are the cases possible. Observe that three cases are possible:
• $a+b = p, a^2 + b^2 - ab =p$
• $a+b =p^2 , a^2 + b^2 - ab = 1$
• $a+b = 1, a^2 + b^2 - ab = p^2$
Now, can you decode these cases and solve the problem like Sherlock?
Observe that a, b are both positive integers. Hence the case: $a+b = 1, a^2 + b^2 - ab = p^2$ is absurd. Let’s concentrate on the other cases one by one. $a+b =p^2 , a^2 + b^2 - ab = 1$ Now,observe this that $a^2 + b^2 - ab = (a-b)^2 + ab = 1$, which is has a solution iff a = b = 1. What about the other case? $a+b = p , a^2 + b^2 - ab = p$ $a+b = p, a^2 + b^2 - ab =p$ Observe a = – b (mod p ) this together with the second equation gives $3a^2 = 0$ (modp). Now p can be 3. For p = 3, Observe that a = 1 and b = 2 is a solution. Now if p is not 3, then p must divide a and b. This implies a + b must be greater than equal to 2p, hence contradiction.

Hence the solutions are a = 1, b =1, p = 2

# 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.

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