Does the equation $k^k+1=a^2$ have any positive integer solutions?

Question

As the title of the question says, I want to know if the equation $k^k+1=a^2$ has any integer positive solutions, (which means k and a are both positive integers.)
I'd really appreciate it if someone could provide the following:

If the given solutions exists, please provide the minimal solution. Are the solutions infinite or finite?
If the given solutions doesn't exist, please provide your proof.

Parcly Taxel · Accepted Answer · 2020-12-17T14:02:24.427

3

There are no positive integer solutions by Mihailescu's theorem.

With a less powerful hammer, Ke Zhao proved the nonexistence of solutions to $x^y+1=a^2$ in 1964. See the last reference in this answer.

edited Dec 17 '20 at 14:02

answered Dec 17 '20 at 13:49

Parcly Taxel

103,344

I did not know that the proof of Catalan's conjecture was so recent. Thanks for the link. – Claude Leibovici Dec 17 '20 at 13:57
@ClaudeLeibovici Actually, a less powerful proof suffices for this question. See edit. – Parcly Taxel Dec 17 '20 at 14:02
Thanks for this one too ! Cheers :-) – Claude Leibovici Dec 17 '20 at 14:05
$$k^k +1=a^2\implies a^2-k^k=1 \implies a = 0 \land k = -1$$ – poetasis Dec 17 '20 at 21:58
To whomever is confused by the comment above; the last step makes no sense. – Servaes Dec 20 '20 at 14:51

score 2 · Answer 2 · answered Dec 17 '20 at 19:15

2

$k^k+1=a^2\Rightarrow k^k=(a-1)(a+1)$

$\gcd(a-1,a+1)=1\text{ or }2$

Case 1: $\gcd(a-1,a+1)=2$. Then $k$ is even, and $k^k$ is a perfect square. The only two squares that differ by $1$ are $0,1$, but $0$ is not a positive integer.

Case 2: $\gcd(a-1,a+1)=1$. Then both $(a-1)$ and $(a+1)$ are perfect $k$th powers that differ by $2$. This can only occur when $k=1$, but $1^1+1=2$ which is not a perfect square.

answered Dec 17 '20 at 19:15

Keith Backman

7,288

This. It's often fun to nuke a mosquito, but when a short and simple argument is available, it should not be left out. With regard to case 1, not only is $0$ ruled out by the positivity requirement, additionally we have $0^0 + 1 = 2$ which is not a square. – Daniel Fischer Dec 17 '20 at 20:09

score 0 · Answer 3 · answered Dec 17 '20 at 14:59

0

$k^k+1$ itself has factors, one of which is $k+1$. These would have to be coprime, or a common divisor of 2, because the common factor between $(a^k+1)$ and $(k+1)$ would divide 2k.

answered Dec 17 '20 at 14:59

wendy.krieger

6,856

Does the equation $k^k+1=a^2$ have any positive integer solutions?

3 Answers3