Prove $5a^2 ≠ 3b^2$ for all non-zero rational numbers $a$ and $b$ In my Discrete Math class, we need to prove there aren't nonzero rational numbers $a$ and $b$ such that $5a^2 = 3b^2$.
I thought about using The Fundamental Theorem of Arithmetic but can't think of a way to apply it. How can I approach this proof in a good manner?
 A: Have you proved that $\sqrt{2}$ is irrational? I believe the proof of this is much the same.
To begin, clear fractions and assume without loss of generality that $a$ and $b$ are integers that are co-prime. (If they share a factor, the square of that factor divides both sides and you can cancel it.)
Next, notice that $3b^2 = 5a^2$ would imply that $3$ must divide $a$. However, this implies that $3^2$ divides $a^2$ and you'll find that $3$ must also divide $b$ a contradiction to our coprime assumption.
Exactly the same could have be done with respect to $5$ instead to get the same result.
A: Let $a=a_1/a_2$, $b=b_1/b_2$, then $5b_2^2a_1^2=3b_1^2a_2^2$.  Now the number of fives in the left side's prime factorisation is odd but even in the right side's factorisation.
A: Yet a complete other solution.  Essentially we want to show that $\sqrt{5/3}$ is irrational.  Now suppose that 
$$\frac{a^2}{b^2}=\frac53\quad\text{with positive  integers $a$, $b$.}$$
It's easy to show that under this assumption the square of
$$\frac{\frac53b-a}{a-b}=
\frac{5b-3a}{3(a-b)}$$
equals $5/3$, too.  Now verify by an easy calculation that both numerator and denominator are positive and
$$5b-3a<a \quad\text{and}\quad3(a-b)<b,$$
which immediately leads to a contradiction.
Motivation: First, let $k$ be natural number which is no square.  Hence there exists a unique natural number $k$ satisfying $n^2<k<(n+1)^2$.
Now $k$ can be approximated be continued fractions $f_1$, $f_2$, and so on.  If $a/b$ is $f_m$, the prior fraction is
$$f_{m-1}=\frac{kb-na}{a-nb}.$$
It's not quite "messy" to show that assuming $(a/b)^2=k$ we get 
$$f_{m-1}^2=k,\quad0<kb-na<a\quad\text{and}\quad0<a-nb<b.$$
The procedure is easily extended if $k$ is a fraction of positive integers, none of which is square (otherwise the positivity will fail), provided $k>1$.
