# In diophantine $3b^2=a^2$ where $a$ and $b$ are coprime, does $3|a$?

Integers $$a$$ and $$b$$ are co-prime and $$3\cdot b^2=a^2$$.

$$3\cdot b^2=a^2$$, implies $$a^2$$ is divisible by 3 since, $$3b^2$$ is divisible by 3.

Is $$a$$ divisible by 3?

• you don't even need $a$ and $b$ to be coprime to deduce. Since $3\mid a^2$, it automatically divides $a$. By the way, I think you're trying to prove that $\sqrt{3}$ is irrational.
– TBTD
May 24 '19 at 14:41

Yes, $$a$$ must be divisible by $$3$$. If the integer $$a=3k+1$$ or $$3k+2$$ for arbitrary $$k$$, (which are the cases in which $$a$$ is not divisible by $$3$$) then the remainder is $$9k^2+6k+1\equiv1\pmod3$$ or $$9k^2+12k+4\equiv1\pmod3$$.

Therefore, $$a^2|3$$ cannot occur if $$a$$ is not divisible by $$3$$, and thus, $$a$$ must be divisible by $$3$$.

However, the equation $$3a^2=b^2$$ cannot be solved for coprime integers, so if you consider that, then it is impossible for $$a|3$$ since there are no $$a$$ that fit the equation.

Yep; in general, if $$p|ab$$ then $$p|a$$ or $$p|b$$ ($$p$$ is a prime). In your case, $$3|a^2$$ so $$3|a$$ or $$3|a$$; thus $$a$$ is divisible by $$3$$

Hint $$:$$ Let $$p$$ be a prime number and $$a,b$$ be positive integers such that $$p \mid ab.$$ Then either $$p \mid a$$ or $$p \mid b.$$

THEOREM: In general, if $$p$$ is prime and $$p|n^2$$, then $$p|n$$, where $$n\inZ$$.

The proof is based on the uniqueness of prime factorisation.

In your case, $$3$$ is a prime and $$a$$ is an integer. Since $$3|a^2$$, this implies that $$3|a$$