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

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

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