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# 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

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