Prove or disprove a claim regarding irrational numbers

Question

I am trying to prove the following claim:

Let $ 0\leq n \in \Bbb Z$ and suppose that there exists a $k \in \Bbb Z$ such that $n=4k+3$.
Prove or disprove: $\sqrt n \notin \Bbb Q$ .

The problem I am having is that I am trying to assume by contradiction that $\sqrt n \in \Bbb Q$ and then I say that there are $a,b \in \Bbb Z$ such that $n=\sqrt {4k+3}=\frac ab$. I finally get to a point where $k=\frac {a^2-3b^2}{4b^2}$. Yet I can't find any $a,b \in \Bbb Z$ that will help me show that the claim is false, nor show a contradiction that will cause the claim to be true.
Any help will be welcomed.

Answer 1

Statement:

Let $a,b,k\in\mathbb Z^{+}$, where $\gcd (a,b)=1$ and if $4k+3=\frac{a^2}{b^2}$, then $b^2=1$ or $b=1$.

Thus we have,

$$\sqrt{4k+3}=a,\thinspace a\in\mathbb Z^{+}$$

and

$$k=\frac{a^2-4+1}{4}=\frac{a^2+1}{4}-1$$

This immediately implies,

$$a=2m-1, \thinspace m\in\mathbb Z^{+}$$

This means,

$$\begin{align}a^2+1&=4(m^2-m)+2\not\equiv 0\thinspace\thinspace\thinspace\text{(mod 4)}.&\end{align}$$

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Conclusion:

We conclude that, there doesn't exist $n=4k+3,\thinspace k\in\mathbb Z^{+}$, such that $\sqrt n\in\mathbb Q^{+}$.