1 plus Square is not Perfect Power

Theorem
The equation:
 * $x^p = y^2 + 1$

has no solution in the integers for $x, y, p > 1$.

Proof
Suppose $p$ is even.

Write $p = 2 k$.

Then:

Since both $y - x^k$ and $y + x^k$ are integers, they must be equal to $\pm 1$.

Summing them up, we have $2 y$ is one of $-2, 0, 2$.

Thus $y$ is one of $-1, 0, 1$, and we ignore these solutions due to our condition $y > 1$.

Now suppose $p$ is odd.

Suppose $y$ is odd.

Then $x^p = y^2 + 1$ is even.

Hence $x$ is even.

Then:

which is a contradiction.

Hence $y$ must be even, and $x$ must be odd.

From Gaussian Integers form Euclidean Domain, we can define greatest common divisors on $\Z \sqbrk i$, and it admits unique factorization.

We factorize $y^2 + 1$:
 * $x^p = y^2 + 1 = \paren {1 + i y} \paren {1 - i y}$

The greatest common divisors of $1 + i y$ and $1 - i y$ must divide their sum and product.

Their sum is $2$ while their product is $y^2 + 1$, which is odd.

Therefore we see that $1 + i y$ and $1 - i y$ are coprime.

From unique factorization we must have that both $1 + i y$ and $1 - i y$ is a product of a unit and a $p$th power.

By Units of Gaussian Integers, the units are $\pm 1$ and $\pm i$.

Hence
 * $\exists u \in \set {\pm 1, \pm i}: \exists \alpha \in \Z \sqbrk i: 1 + i y = u \alpha^p, 1 - i y = \bar u \bar \alpha^p$

Since $p$ is odd:
 * $1^p = 1$
 * $\paren {-1}^p = -1$
 * $i^p = \pm i$
 * $\paren {-i}^p = -i^p = \mp i$

therefore there is some unit $u' \in \set {\pm 1, \pm i}$ such that $u'^p = u$.

By writing $\beta = u' \alpha$:
 * $1 + i y = u'^p \alpha^p = \beta^p, 1 - i y = \bar \beta^p$

Write $\beta = a + i b$, where $a, b \in \Z$.

By Sum of Two Odd Powers:
 * $2 a = \beta + \bar \beta \divides \beta^p + \bar \beta^p = 2$

this gives $a = \pm 1$.

We also have:

since $1 + y^2$ is odd, $b$ must be even.

Hence:

In particular, comparing real parts gives $1 \equiv a^p \pmod 4$.

Since $p$ is odd, we have $a = 1$.

Now we have:

The summands on the may not be an integer, but if we can show:
 * In canonical form, the numerator of each summand is even

then the equation is never satisfied.

This is because the sum of all the terms will be a rational number with even numerator and odd denominator, which cannot equal to $1$.

Since $2 k + 1$ is always odd and $\paren {-1}^k \dbinom {p - 2} {2 k - 2}$ is always an integer, we only need to check $\dfrac {b^{2 k - 2} } k$.

Since $b$ is even:
 * $2^{2 k - 2} \divides b^{2 k - 2}$

But we have:

Hence the largest power of $2$ that divides $k$ is less than $2^{2 k - 2}$.

Therefore the numerator of $\dfrac {b^{2 k - 2} } k$ is even.

And thus all the equations above are never satisfied.

So our original equation:
 * $x^p = y^2 + 1$

has no solution in the integers for $x, y, p > 1$.