Is $x^2+x-1$ squarefree? I noticed that the expression $f(x) = x^2+x-1$ seemed to be squarefree for every positive integer value of $x$. Equivalently, there does not exist a prime $p$ for any $x \in \mathbb{Z}^+$ such that $$\dfrac{x^2+x-1}{p^2}$$ is an integer. Is this true and if so how do we prove it?
 A: Its not square free. Following are few examples:
x = 36, f(x) = 1331 and is divisible by  11*11
x = 42, f(x) = 1805 and is divisible by  19*19
x = 84, f(x) = 7139 and is divisible by  11*11
x = 157, f(x) = 24805 and is divisible by  11*11
x = 198, f(x) = 39401 and is divisible by  31*31
x = 205, f(x) = 42229 and is divisible by  11*11
x = 226, f(x) = 51301 and is divisible by  29*29
x = 278, f(x) = 77561 and is divisible by  11*11
x = 318, f(x) = 101441 and is divisible by  19*19 
A: It is quite easy to see that this proposition cannot be true from a purely theoretical perspective. Fix a prime $ p \neq 2, 5 $, and that is $ 1 $ or $ 4 $ modulo $ 5 $, for example $ p = 11 $. (The reason for this choice will become clear shortly.) Working in $ \mathbf Z/p^2 \mathbf Z $, we can write
$$ X^2 + X - 1 = 0 $$
$$ \left( X + \frac{1}{2} \right)^2 = \frac{5}{4} $$
where the inverses make sense since $ p \neq 2 $. On the other hand, squares in $ \mathbf Z/p \mathbf Z $ lift to squares in $ \mathbf Z_p $ (the p-adic integers) by Hensel's lemma, therefore $ 5 $ is a quadratic residue modulo $ p^2 $ iff it is a quadratic residue modulo $ p $. By quadratic reciprocity, this is true if and only if $ p $ is $ 1 $ or $ 4 $ modulo $ 5 $. Hence, we may take square roots on both sides and find the solution
$$ X = \frac{\pm \sqrt{5} - 1}{2} $$
For instance, $ 48^2 = 5 $ modulo $ 11^2 $, so for $ X = 47/2 = 47 \cdot 61 = 84 $ modulo $ 121 $, the given expression is divisible by $ 121 $. This produces infinitely many such examples. Note that exactly the same method yields that if $ p \equiv 2, 3 \pmod{5} $ then $ p $ cannot divide $ X^2 + X - 1 $ for any value of $ X $, so the examples found this way are the only ones, a result which agrees with the computational answer provided by maverick.
