Prove that no integer of the form $7k+3 (k \in \mathbb Z) $ is a perfect square.

Question

What I have is this, and I would like to make sure that I did it correctly.

Suppose an integer of form $7k + r$, where $k \in \mathbb Z$ and $0 < r < 7$ is a perfect square. Then $7q+r =0$ ore $7q+r =1$. Therefore $r = 0$, and $r = 1$.

Answer 1

I think I didn't actually understand your solution.

The easiest one I can think of is taking $(7k + i)^2 (mod 7) = i^2 (mod 7)$ for $i = 0,...,6$. You get the values $0,1,4,2,2,4,1$, but never $3$, so a perfect square can never be of that form.

Answer 2

Use quadratic reciprocity: $$ \left(\frac{3}{7}\right)=\left(\frac{7}{3}\right)(-1)^{\frac{3-1}{2}\frac{7-1}{2}}=-1. $$

Answer 3

Think of this in terms of quadratic residues. You can restate the question as "what square numbers have residues of 3 (mod 7)"? Then we have

$$a^2 \equiv 3~(\text{mod}~7)$$

Now, $$1^2 \equiv 1~(\text{mod}~7)$$ $$2^2 \equiv 4~(\text{mod}~7)$$ $$3^2 \equiv 2~(\text{mod}~7)$$ $$4^2 \equiv 2~(\text{mod}~7)$$ $$5^2 \equiv 4~(\text{mod}~7)$$ $$6^2 \equiv 1~(\text{mod}~7)$$

And since we never achieve the desired congruence of 3, no such integer k exists that would make $7k+3$ a perfect square.