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## Homework Statement

Show that if

*n*and

^{2}+ m*n*are perfect squares then m is divisible by 24.

^{2}- m## Homework Equations

This problem comes from Larson's

*Problem-solving Through Problems*in the section on modular arithmetic. It is the third part of a four-part problem, with the previous two parts of the problem being to prove that if

*x*has a solution in integers x, y, and z then one of them is divisible by seven, and the second part to prove that if

^{3}+ y^{3}= z^{3}*2*is prime then n is congruent to 3 mod6. Since they're part of the same problem I assume that they are somehow relevant.

^{n}+ n^{2}## The Attempt at a Solution

I want to show that m is divisible by both 3 and 8. By the Chinese Remainder Theorem, I can set up the following:

m ≡ a(mod3)

m ≡ b(mod8)

And I assume that what I want to do is show that the two expressions can only be perfect squares if a = b = 0.

I don't know if any of the following actually brings me any closer to a solution, but I feel like they might be pointing in the right direction:

I can set up the following relations:

*n*

^{2}*+ m ≡ n*(mod3) +

^{2}*a(mod3)*

*n*

^{2}*+ m ≡ n*(mod8) +

^{2}*b(mod8)*

n

n

^{2}*- m ≡ n*(mod3) -

^{2}*a(mod3)*

n

n

^{2}*- m ≡ n*(mod8) -

^{2}*b(mod8)*

And further I can mod n out so that

*n*(mod3) +

^{2}*a(mod3)*≡

*[a(mod3)*](

*modn*), etc

I can also multiply

*n*and

^{2}+ m*n*to get

^{2}- m*n*, which gets

^{4}- m^{2}*n*-

^{4}*m*

^{2}≡ (-a^{2}mod3)(modn)*n*- m^{4}^{2}≡ (-b^{2}mod8)(modn)All of which I think might be useful in establishing a = b = 0, but I can't think of how to get there.