# $a^2+b=b^{1999}$ how many integer pairs $(a,b)$ are there that satisfy this equation.

How many integer pairs $(a,b)$ are there such that $a^2+b=b^{1999}$

this is the original text unfortunately it doesn't give much information. I have found one possible pair but I can't prove that there are no more solutions, is this about something like deciding which side is odd and even?

My attempts;

$a^2=b(b^{1998}-1)$ now we can weakly say that $b=b^{1998}-1$ however that does not give me a solution at all another attempt was to guess $(a,b)$ and my first guess would be $a=0$ and $b=\pm1$ but I can't get to a rather formal solution from here. What are your suggestions?

• Where is this problem from?
– user472341
Sep 12, 2017 at 11:21
• An olympiad preparation text book. Sep 12, 2017 at 11:22

$$a^2=b(b^{1998}-1)$$
As $(b,b^{1998}-1)=1$ if $a\ne0$ both have to be perfect square
So, we need $$b^{1998}-1=d^2\iff(b^{999}+d)(b^{999}-d)=1$$ for some integer $d$
• Which means $b=\pm1$ thank you:)) Sep 12, 2017 at 11:24
• It also implies $b=0$ Sep 12, 2017 at 12:17
Since $\gcd(b,b^{1998}-1)=1$ we know that $b$ is a square. Thus we want $k^2=b^{1998}-1$. If $|b|>1$ then $(b^{999})^2>b^{1998}-1>(b^{999}-1)^2$. Thus from square bounding $|b|\leq1$. Now case bash the remaining cases.