# How prove the $\sqrt{n^2+1}$ is irrational?

I want to know how can we prove $$\sqrt{n^2+1}$$ is irrational. I know that when I use to find the square root of something by the long division method as suggested in the below image( it about some other expression and does not demonstrate my question but I attached the image to show the method of long division to find square roots.) and we get a remainder them it is irrational but I am not able to know why getting a remainder means irrational.

I want it to be opened because the duplicate suggested what I have already answered but not what I want to know, I want to know how can prove that $$\sqrt{n^2+1}$$ is irrational by the long division method used to find square roots.

I am talking about this me thod. Here it is used to find $$\sqrt{4x^4-28x^3+37x^2+42x+9}$$ This is just an example to show the method. My question is above the picture.

• What do you mean by "finding the square root by long division"? And what does it mean that $x^2+3x+1$ is the quotient and $-1$ is the remainder?
– user562983
Commented Sep 23, 2021 at 8:02
• @SaucyO'Path It means the poly is $(x^2+3x+1)^2 -1$ - I suppose one can do the poly analogue of the sqrt process for numbers which is a sort of analogue of long division? (!) Commented Sep 23, 2021 at 8:05
• @boojum I am not sure about that, it could for a given $x$ be a square without there being a polynomial answer. Commented Sep 23, 2021 at 8:07
• @boojum $x(x+1)(x+6)(x+25)$ is a square for $x=2$, but I assure you that it isn't the square of a polynomial. You can do something of sort if you find a polynomial $g$ such that $p(x)-g^2(x)$ isn't equal to the dfference of two squares for any $x$, but that's more of a circumstantial occurrence than a method (as you can see by the tautological formulation).
– user562983
Commented Sep 23, 2021 at 8:15
• @SaucyO'Path By Long division I mean one of the methods used to find square root and qoutionent and the remainder obtained in it. Commented Sep 23, 2021 at 9:18

Suppose $$\sqrt{n^2 + 1}$$ is rational (in lowest terms). $$\sqrt{n^2 + 1} = \frac{a}{b} \text{,}$$ where we may require $$a,b \in \Bbb{Z}$$, $$b > 0$$, and $$\gcd(a,b) = 1$$. We have $$n^2 + 1 = \frac{a^2}{b^2} \text{.}$$ The left-hand side is an integer, so the right-hand side is an integer. Since $$\gcd(a,b) = 1$$, this requires $$b = 1$$. Therefore, $$n^2 + 1 = a^2 \text{.}$$ But then \begin{align*} 1 &= a^2 - n^2 \\ &= (a+n)(a-n) \text{.} \end{align*} If $$n = 0$$, this is a factorization of $$1=a^2$$ and so $$a =-1$$ or $$a=1$$ is a solution. If $$n \neq 0$$, this is a factorization of $$1$$ into two distinct integers, which is impossible. (The only divisors of $$1$$ are $$-1$$ and $$1$$ and we have already seen how those factors produce $$1$$.)
Therefore, either $$n = 0$$ or $$\sqrt{n^2+1}$$ is irrational.
• @EricTowers Your answer points out the necessity for $n≠0$ for the square root to be irrational. And also I am interested by the method I used for square roots for proof , if you can. Commented Sep 24, 2021 at 3:31
• @MohdSaad : No problem. In the first step, you find the leading term in $\sqrt{n^2+1}$ is $n$. Then in the second step, $n$ doesn't divide the remainder from the first step, $1$. Commented Sep 24, 2021 at 12:08