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.
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4$\begingroup$ consider $f(x)=x^b-b^x-3$ and use dervative. $\endgroup$– hamam_AbdallahOct 22, 2016 at 14:56
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1 Answer
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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 $(4,1)$
answered Oct 22, 2016 at 15:10
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$\begingroup$ Should use $\ge$ (dollar sign, \ge, dollar sign) instead of >. $\endgroup$ Oct 22, 2016 at 15:15
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1$\begingroup$ @OscarLanzi yeah now I'll edit, thank you. $\endgroup$ Oct 22, 2016 at 15:18