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.

  • $\begingroup$ $k$ is an integer. So that, you don't need $b$. Thus, $b=1$. $\endgroup$ Oct 24 at 20:24
  • $\begingroup$ Try to solve “If $a^2-3b^2$ is divisible by $4,$ then $a,b$ are both even.” $\endgroup$ Oct 24 at 20:25
  • $\begingroup$ @lonestudent What? That’s a jump. $\endgroup$ Oct 24 at 20:27
  • $\begingroup$ @ThomasAndrews Am I thinking wrong?.. $\endgroup$ Oct 24 at 20:29
  • $\begingroup$ Well, $\frac A{4b^2}$ can be an integer, without $b=1.$ The question is what about $a^2-3b^2$ let’s us say this particular $b$ must be $1.$ Hence I called it “quite a jump” rather than “wrong.” @lonestudent $\endgroup$ Oct 24 at 20:39


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,

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



This immediately implies,

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

This means,

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


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

  • $\begingroup$ I don't understand why $k \in \Bbb Z$ implies that $\sqrt {4k+3} \in \Bbb Z$ For example for $k=1$ we have that $a=\sqrt 7 \notin \Bbb Z$ $\endgroup$ Oct 24 at 20:39
  • $\begingroup$ I understand but no one said that $\sqrt n$ is an integer, thus I don't know why you assume that ${4k+3}$ is a perfect square. $\endgroup$ Oct 24 at 20:57
  • $\begingroup$ @Yuval I expanded my answer. Now, everything is clear. $\endgroup$ Oct 24 at 21:15

The square of an even integer is $4k$, the square of an odd integer is $8k+1$. $4k+3$ is never the square of an integer, neither is $4k+2$ nor $8k+5$. So the square root of $n$ is not an integer.

Now you need to remember the well known proof that the square root of 2 is irrational; that proof can be adapted to show that the square root of any integer is either an integer or irrational.


It is given that $n=4k+3$, so I think you mean to say that there exist $a,b\in\Bbb{Z}$ such that $$\sqrt{n}=\sqrt{4k+3}=\tfrac ab,$$ in the hopes of reaching a contradiction. Indeed some algebra then leads to $$k=\frac{a^2-3b^2}{4b^2},$$ which means that $4b^2$ should divide $a^2-3b^2$ because $k$ is an integer. In particular $4$ should divide $a^2-3b^2$. This implies that $a$ and $b$ are both even [prove this!], say $a=2A$ and $b=2B$. Plugging this in then gives $$k=\frac{(2A)^2-3(2B)^2}{4(2B)^2}=\frac{4A^2-12B^2}{16B^2}=\frac{A^2-3B^2}{4B^2},$$ and so by the same argument $A$ and $B$ are again both even. Can you see the contradiction from here?

  • $\begingroup$ I understand that this process can go on and on and reach the same point every time, but how can I reach a contradiction from here? $\endgroup$ Oct 24 at 20:43
  • $\begingroup$ There are two ways to proceed. One way is to conclude that both $a$ and $b$ are integers that can be divided by $2$ indefinitely, which implies $a=b=0$, a contradiction. The other way is to start with a reduced fraction $\tfrac{a}{b}$, so that $a$ and $b$ are coprime. Then the conclusion that both are even is already a contradiction. $\endgroup$
    – Servaes
    Oct 24 at 20:44

Note the if $q$ is rational but not integral then $q^2$ will also not be integral. So you only need to prove that $n$ is not a power of an integer.

Note that if $m=2l$ then $m^2=4l^2$, if $m=4l+1$ then $m^2=16l^2+4l+1$ and if $m=4l+3$ then $m^2 = 16l^2+12l + 8 +1$. So for any $m$ we have that $m^2$ is either of the form $4r$ or of the form $4r+1$.

In different words: Modulo $4$ we have $0^2=0,1^2=1,2^2=0,3^2=1$.

So any number of the form $4k+3$ is not a square of an integer.


$\sqrt n=\sqrt{4k+3}$ is a solution of the equation $$x^2 -(4k+3)=0$$

Now by Rational Zeros Theorem $\biggr($Which states that : $x=p/q$ is a Rational Number satisfying the polynomial equation $\sum_{r=0}^n a_rx^r $ then $q(\neq0)|a_n$ and $p|a_0$ and gcd(p,q)=1 $\biggr)$ , the only Rational roots of the equation are $±1,±(4k+3),±n|(4k+3) \forall k\in \Bbb Z^+$ where $n \in \Bbb Z^+$. But $\sqrt{4k+3}$ is neither of them . So $\sqrt n=\sqrt{4k+3}$ is irrational.


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