Find positive integers
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Jarif Mahbub Srabon
- Posts:15
- Joined:Thu May 31, 2012 11:39 pm
- Location:Mymensingh, Bangladesh
Find out positive integers for $a,b$ as if $a^b+b^a=999$
Last edited by nayel on Sun Oct 21, 2012 6:32 pm, edited 1 time in total.
Reason: Irrelevant subject
Reason: Irrelevant subject
Re: Find positive integers
$(a,b)=(998,1),(1,998)$. 
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sakibtanvir
- Posts:188
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Re: Find positive integers
Yes,the only solutions are $(a,b)=(1,998),(998,1)$.
It is easy to notice one of the variables $(a,b)$ is odd and another is even.WLOG let us assume $b=2k$(even).
Case1:When $a \leq 1$,the only solutions found by a little investigation is $(a,b)=(1,998),(998,1)$.
Case2:When $a,b >1$,$a^{2k}+(2k)^{a}=(a^{k})^{2}+2^{a}k^{a} \equiv 1 (mod4)$, A contradiction.
It is easy to notice one of the variables $(a,b)$ is odd and another is even.WLOG let us assume $b=2k$(even).
Case1:When $a \leq 1$,the only solutions found by a little investigation is $(a,b)=(1,998),(998,1)$.
Case2:When $a,b >1$,$a^{2k}+(2k)^{a}=(a^{k})^{2}+2^{a}k^{a} \equiv 1 (mod4)$, A contradiction.
An amount of certain opposition is a great help to a man.Kites rise against,not with,the wind.
Re:Find positive integers
Very nice solution!sakibtanvir wrote:Yes,the only solutions are $(a,b)=(1,998),(998,1)$.
It is easy to notice one of the variables $(a,b)$ is odd and another is even.WLOG let us assume $b=2k$(even).
Case1:When $a \leq 1$,the only solutions found by a little investigation is $(a,b)=(1,998),(998,1)$.
Case2:When $a,b >1$,$a^{2k}+(2k)^{a}=(a^{k})^{2}+2^{a}k^{a} \equiv 1 (mod4)$, A contradiction.
"Everything should be made as simple as possible, but not simpler." - Albert Einstein
Re: Find positive integers
প্রবলেম টা দেখতে কঠিন, কিন্তু সল্যুশন সহজ আর মজার।