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If $x$, $y$ and z are positive integers, I want to solve $$2015+x^2=3^y4^z$$

What I tried:

I found that $x \equiv 1$ (mod $24$). So there exist an integer $p$ such that $24p+1=x$. Replacing that on the equation, we have $$48(12p^2+p+42)=3^y4^z$$

Behrouz Maleki
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    $(x,y,z)=(17,2,4)$ is a solution. It's not true that $x\equiv 1\pmod{24}$ (we have $17\not\equiv 1\pmod{24}$). – user236182 Aug 08 '16 at 08:41
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    If $y=2r$ is even, you have $2015=(3^r2^z)^2-x^2$, and it's easy to find all the ways to write 2015 as a difference of two squares and to figure out which ones, if any, involve a number of the form $3^r2^z$. If $y=2s+1$ is odd, then you have $3u^2-x^2=2015$ with $u=3^s2^z$, to which you may be able to apply techniques for Pell equations. – Gerry Myerson Aug 08 '16 at 09:13
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    No need to go for Pell. This thing has no solutions modulo 25, which leaves us with the first option alone, which gives the already mentioned solution. – Ivan Neretin Aug 08 '16 at 09:13
  • @IvanNeretin No solutions $\pmod {25}$ would imply the solution $(17, 2, 4)$ cannot exist either. So obviously you're wrong. – Macavity Aug 08 '16 at 11:05
  • $3^y4^z$ is either a square (option 1) or three times a square (option 2). I was talking about option 2. – Ivan Neretin Aug 08 '16 at 11:13

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