7
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ Where is this problem from? $\endgroup$
    – user472341
    Sep 12, 2017 at 11:21
  • $\begingroup$ An olympiad preparation text book. $\endgroup$ Sep 12, 2017 at 11:22

2 Answers 2

7
$\begingroup$

$$a^2=b(b^{1998}-1)$$

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 $d$

$\endgroup$
2
  • $\begingroup$ Which means $b=\pm1$ thank you:)) $\endgroup$ Sep 12, 2017 at 11:24
  • $\begingroup$ It also implies $b=0$ $\endgroup$
    – InsideOut
    Sep 12, 2017 at 12:17
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.