First Post of the Year(A Very easy problem)

For discussing Olympiad Level Number Theory problems
tanmoy
Posts: 289
Joined: Fri Oct 18, 2013 11:56 pm
Location: Rangpur,Bangladesh

First Post of the Year(A Very easy problem)

Unread post by tanmoy » Fri Jan 01, 2016 1:40 pm

Solve the equation $x^{2}+y^{4}+1=6^{z}$ in the set of integers.

[$\text{Source:2015 JBMO TST - Macedonia, Problem 1}$]
"Questions we can't answer are far better than answers we can't question"

rah4927
Posts: 108
Joined: Sat Feb 07, 2015 9:47 pm

Re: First Post of the Year(A Very easy problem)

Unread post by rah4927 » Sat Jan 02, 2016 10:32 pm

First note that $z$ is necessarily non-negative (otherwise the left side can't be an integer).

First assume $z>0$

We have that $x^2+y^4$ is odd, which implies that one of $x,y$ is even while the other is odd. Suppose $x$ is even. Then taking both sides of the equation modulo $4$, we have that $2\equiv 2^z$ but that's possible only if $z=1$. On the other hand, if $x$ is odd, then again taking mod $4$, we get $2\equiv 2^z$ and so $z$ has to be $1$.

Now check the remaining cases by hand to find the solutions(s).

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