show that the square root of (7n+3) is irrational (n ∈N∗)

Question

somebody gave me this answer to the question in the title:

A2A: Observe that the remainder you get when you divide any square by 7 is never 3. You only need to observe it for the squares of the first 7 integers.

You also have to know that √m is rational if and only if m is a perfect square.

why does the remainder need to be 3 for it to be a perfect square?

In: 0

Answer 1

Remainder does not need to be 3 for it to be a perfect square – the opposite, actually: perfect squares **never** have remainder 3 (when divided by 7).

It’s just that the number `7n + 3` *does* have a remainder 3 (`7n` has remainder 0, `3` has remainder 3, so `7n + 3` has remainder 0+3).

But, because perfect squares never have remainder 3 (div. 7), `7n+3` is not a perfect square.

But only perfect squares have rational roots, so √(7n+3) is not rational.

Therefore, it is irrational.