Finding Possible Solutions for a^3 = 5b^3: Considerations and Limitations

[recon]

I've been asked to find all possible pairs (a,b) of integers such that a^3 = 5b^3.

I started by considering the cases where a and b share a common factor, so that a = kl and b = km. But this can be reduced to the form l^3 = 5m^3, so we can assume WLOG that a and b are coprime.

If this is the case, then 5 \mid a^3, so 5 \mid a. Let a = 5x.

(5x)^3 = 5b^3
125x^3 = 5b^3
25x^3 = b^3

So 25 \mid b^3 and 5 \mid b. Let b = 5y.

This contradicts our earlier assumption that both a and b are coprime, and hence no solution exists.

Have I made any mistakes?

 

[Data]

Well, there's one solution!

0^3 = 5(0)^3

Other than that, your answer is fine. You just need to modify it for the zero case (if a=b=0 then one of your steps doesn't work).

 

[recon]

Ah, of course...there's always that pesky number called zero!

Thanks for pointing that out Data!