# How to show that $(a/b)^2$ in this situation cannot be an integer [duplicate]

Let $$a$$,$$b$$ be relatively prime integers which are both greater than or equal to $$􏰀2$$ . Show that $$(a/b)^2$$ cannot be an integer.

I have tried to use contradiction to prove this result, but I am not sure that my idea is correct or not. Below is my idea (not a complete proof).

Suppose $$(a/b)^2$$ be an integer. We have $$a/b$$ is an integer. Then, we have $$a=bk$$ where $$k$$ is an integer since $$b$$ divides $$a$$. Since $$a,b$$ are relatively prime, therefore $$b$$ must be $$1$$. Hence, contradiction arises. I am not sure that my idea is correct or not, could anyone help me to check it? If my idea is wrong, could you give me a idea to do this question?

• You already know that $a/b$ is not an integer. It seems to me what is really being asked is to show that if a rational number is not an integer, then neither is its square. – MPW Mar 18 '20 at 15:11
• You have to exclude the case that $(a/b)^2=m$ is not a perfect square. – Maik Pickl Mar 18 '20 at 15:11
• Your very first line is not quite right. If $(a/b)^2$ is integer, then it does not follow that $a/b$ is integer. Say, what if $(a/b)^2 = 2$? – user726394767 Mar 18 '20 at 15:11