5
$\begingroup$

This question is given in AMTI second level contest.

Find all pairs of naturals $(a,b)$ such that $a^b-b^a=3$.

My try: One such pair is $(4,1)$. Are there any other pairs? Please help me.

$\endgroup$
  • 4
    $\begingroup$ consider $f(x)=x^b-b^x-3$ and use dervative. $\endgroup$ – hamam_Abdallah Oct 22 '16 at 14:56
  • $\begingroup$ I didn't understand you sir. $\endgroup$ – sai saandeep Oct 22 '16 at 15:32
3
$\begingroup$

Taking mod 2 we can easily see that exactly one of them is even .
Case 1. $a=2m$:
$$(2m)^b-b^{2m}=3$$
If $b\geq3$, $(2m)^b\equiv 0\pmod 8$ so
$$-b^{2m}\equiv 3\pmod 8$$ $$b^{2m}\equiv 5\pmod 8$$ but no square is 5 mod 8,so no solution, so $b=1$
$$2m-1=3$$ $$m=2,a=4$$ Case 1. $b=2m$:
$$(a)^{2m}-(2m)^{a}=3$$
If $a\geq3$, $(2m)^a\equiv 0\pmod 8$ so
$$a^{2m}\equiv 3\pmod 8$$ but no square is 3 mod 8,so no solution, so $a=1$
$$1-2m=3$$ but no natural solution
The only solution is $(4,1)$

$\endgroup$
  • $\begingroup$ Should use $\ge$ (dollar sign, \ge, dollar sign) instead of >. $\endgroup$ – Oscar Lanzi Oct 22 '16 at 15:15
  • 1
    $\begingroup$ @OscarLanzi yeah now I'll edit, thank you. $\endgroup$ – arberavdullahu Oct 22 '16 at 15:18

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.