Tighten it up

$x^{x}+1=y^{2}$

Now , $x=-1$ is the $ONLY$ possible case.

$-1^{-1}+1=y^{2}$

$\frac{1}{-1} +1=y^{2}$

$-1+1=y^{2}$

$0=y^{2}$

Therefore ,

$y=0$

Now , Only one pair is possible .

$\boxed{(-1,0)}$

• For $x < -1$ the value of $y$ gets out of the range of the integers.
• For $x = -1$ we get the pair $(x, y) = (-1, 0)$ .
• For $x = 0$ we get no answer since $0^0$ is indeterminate.
• For $x$ being an even number, let $x = 2l$ , for $l$ an integer: $x^{2l} + 1 = y^2 \\ \Rightarrow (x^l)^2 + 1^2 = y^2 \leftarrow\big(k\cdot2mn\big)^2+\big(k(m^2-n^2)\big)^2=\big(k(m^2+n^2)\big)^2$

There isn't any non-primary or primary pythagorean triplet constituted of $1$ since: $k\cdot 2mn \neq 1$ and $k\cdot(m^2 - n^2) \neq 1$ for $k, m, n$ integers.

• For $x$ being an odd number, let $x = 2a + 1$ for $a$ an integer. Hence $x^x$ is odd too and we can get this by the binomial theorem: $\\ x^x= (2a+1)^{2a+1} \\ =\bigg(\binom{2a+1}{0}\cdot (2a)^{2a+1}\cdot 1^0\bigg)+\ldots+\bigg(\binom{2a+1}{2a+1}\cdot (2a)^{0}\cdot 1^{2a+1}\bigg) \\ =\bigg(\binom{2a+1}{0}\cdot (2a)^{2a+1}\cdot 1^0\bigg)+\ldots+\bigg(\binom{2a+1}{2a}\cdot (2a)^{1}\cdot1^{2a}\bigg)+1$

Hence: $x^x + 1 = y^2 \\ \Rightarrow x^x = y^2 - 1 \\ \Rightarrow x^x = (y - 1)\cdot (y + 1)$

Since $x^x$ is odd, then both $(y - 1)$ and $(y + 1)$ must be consecutive odds and coprime. And we conclude this as follows:

Let $p$ be the gcd of $(y - 1)$ and $(y + 1)$ , then $p | (y + 1) - (y - 1) \Rightarrow p | 2 \Rightarrow p \in \big\{1 ,2\big\}$ . But $(y - 1)$ and $(y + 1)$ are odds, so $p = 1 \Rightarrow \gcd\Big((y - 1),(y + 1)\Big) = 1$ .

Let $x = q_1\cdot q_2\cdot q_3\ldots q_n$ , where $q_i (1\leq i \leq n)$ represents every single prime factor of $x$ at its greatest exponent. Since $(y + 1)$ and $(y - 1)$ are relatively prime, then they must both be an integer raised to the power of $x$ , such that $(y - 1) = a^x$ and $(y + 1) = b^x$ , where $a$ and $b$ are integers (and $a\cdot b = x$ ). However for $x \geq 3$ , $a^x$ and $b^x$ cannot differ by $2$ and for $x = 1$ or $x = 2$ we get no integral solution for $y$ .

x^x + 1=y^2..

For solving this problem, we need a little imagination.

lets deal with positive integer value of x first-

CASE 1- IF X IS EVEN- when x is even, x^x will itself be a perfect square. So on adding 1 to it, we'll not get a perfect square.so an integeral value of y cant be obtained. So x cant be a positive even no. EX.- 4^4 is a perfect square but 4^4+1 is not.

CASE 2-IF X IS ODD- when x is odd, x^x(x-1) is the only factor which can be added to x^x to make it a perfect square. So on adding 1 to x^x will not make it a perfect square.

x^x + x^x(x-1) = x^x(x) = x^(x+1)...since x+1 is even, this term is a perfect square.

EX.- 7^7 + 7^7(7-1) = 7^8 which is a perfect square. while 7^7 + 1 is not.

So above two logics illustrate that x cant be a positive no. and not even 0 as 0^0 is not defined.

IF X IS A NEGATIVE INTEGER OTHER THAN -1 - if x is < -1, then the term x^x will become a fraction...so y cant be obtained as integer...

The only possible value for which y is integer is -1. so y becomes 0.