# Does a solution exist?

1. Apr 7, 2005

### 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.

2. Apr 7, 2005

### 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).

Last edited: Apr 7, 2005
3. Apr 7, 2005

### recon

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

Thanks for pointing that out Data!