can a number of the form $x^2 + 1 $ be a square number?

Question

I have been trying to prove that $x^2 + 1 $ is not a perfect square (other than $0^2 +1^2=1^2$). I'm stuck and can't move forward.

The thing I have tried so is to relate the problem to a hyperbola and find an integer solution for both $x$ and $y$ when $a=b=1$. The pell's equation came up in my search, but I don't understand it fully.

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^{Note: I was in a confused state and @CoolHandLouis' visual answer cleared my muddled mind, so I selected that answer. In that way, his answer was very helpful to me. @Alessandro's proof is clear to me now and if I could accept two answers, I would accepted that one too. Thanks to everyone for helping!}

Answer 1

$(n+1)^2-n^2=2n+1$, that is, the difference of consecutive squares is the $n$-th odd number.

Since 1 is the first odd number it is the difference of the second and the first square: $0^2+1=1^2$

Answer 2

If $x>1$, then $x^2+1$ cannot be a perfect square because $x^2<x^2+1<(x+1)^2$.

In other words, $x^2+1$ lies between the two consecutive perfect squares $x^2$ and $(x+1)^2$.

Answer 3

If $x,y$ are integers with $x^2+1=y^2$, then $1=y^2-x^2=(y+x)(y-x)$. The only integer factorizations of $1$ are $1=1\cdot 1=(-1)\cdot (-1)$, hence $x=0$ and $y=\pm1$.

Answer 4

We want to prove $x^2 + 1$ can never be a perfect square.

Let

• $f(x) = x^2$

  Then,

  $f(x)$     $<$     $f(x) + 1$     $<$     $f(x+1)$

  $x^2$         $<$        $x^2 + 1$     $<$      $x^2 + 2x + 1$        (for all $x > 0$).

Therefore, $x^2 + 1$ cannot be a perfect square (except $x = 0$) because it will always be greater than the prior perfect square and less than the next perfect square.

The following table illustrates this. Note that $f(x)$ is the set of all perfect squares:

x    f(x)=x^2       x^2+1       f(x+1)    
0        0            1            1     
1        1            2            4        
2        4            5            9        
3        9           10           16        
4       16           17           25        

Answer 5

There's a more general result called Catalan's conjecture (now proved).

$$x^y+1$$ is never a perfect power for any natural $x,y>1$ (except for $(x,y)=(2,3)$).

Answer 6

Given an integer $n$, $n^2 = n \times n$. That's obvious enough.

But what's $n^2 - (n - 1)^2$? As it turns out, it's $2n - 1$ (let me know if you want me to elaborate on that).

Then we're looking for solutions to $n^2 = x^2 + 1$ where $x$ is also an integer. Since both $n$ and $x$ are integers, $n^2$ and $x^2$ must be consecutive perfect squares. This leads to $$n^2 - x^2 = 2n - 1 = 1.$$ The only possible solution with $n$ positive is $n = 1$, so that $2n = 2$ and $2 - 1 = 1$.

What if we allow negative integers? There is one other solution: $n = 0$, $x = -1$.

There is also something called imaginary numbers. There might be a solution among them, but I only barely understand that concept, so I wouldn't even be able to begin looking for a solution among those.