# 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?

• Hint: You can suppose $a$ and $b$ are non-zero integers. – Jakobian Jul 3 '19 at 14:23
• Hint: Write $a=\frac{p}{q}$ and $b = \frac{m}{n}$ to reduce the problem to when $a$ and $b$ are integers. – AHusain Jul 3 '19 at 14:23

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.

• How do you conclude that 3 must divide a? – F. Zer Jul 3 '19 at 14:29
• @F.Zer Since we require everything to be an integer, the left-hand side is an integer and so $3$ divides $5a^2$. However, $3$ certainly doesn't divide $5$ and so it must divide $a^2$. i.e. It divides $a$. – Laarz Jul 3 '19 at 14:36
• Thank you. I understood, now. So, $3^2$ divides $a^2$. How do you know 3 must also divide $b$? – F. Zer Jul 3 '19 at 16:30
• Since $3^2$ divides $5a^2$, $3^2$ must also divide $3b^2$, so $3$ must divide $b^2$. – Martin Kochanski Jul 3 '19 at 17:36

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.

• Worth emphasis: such sketches implicitly use the Fundamental Theorem of Arithmetic (or a simpler form that every integer can be written uniquely in the form $\, 5^{\large k} n\,$ where $\,5\nmid n).\,$ To obtain a rigorous proof it is essential to be explicit about invoking these basic Theorems (which are crucial to the proof). – Bill Dubuque Jul 3 '19 at 19:12
• Above I meant "every nonzero integer". – Bill Dubuque Jul 4 '19 at 0:16
• See my alternative solution, please. – Michael Hoppe Jul 5 '19 at 16:19

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 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.

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

The procedure is easily extended if $$k$$ is a fraction of positive integers, none of which is square (otherwise the positivity will fail), provided $$k>1$$.

• +1 I like this answer, and I agree with the statement that $(5/3b-a)/(a-b)$ squares to $5/3$, but the proof that I can see relies on some messy algebra. Your exposition also leaves open the question of how you came up with this proof, which would be nice. – Cheerful Parsnip Jul 16 '19 at 18:33
• See my expanded answer, please. – Michael Hoppe Jul 17 '19 at 11:56