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Can we find solutions of Diophantine equations of the form :

$$(2^n)^x + p^y = z^2 $$

where $k, x, y, z$ and $n$ are positive integers.

-Richard Simson

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What does p belong to? – Tomarinator Mar 19 '12 at 11:07

Some easily-spotted solutions:

  1. $p = 0$, $nx$ is even, $z$ is a square root of $(2^{nx})$.
  2. $x = y = 2$, $(2^{n},p,z)$ is a Pythagorean triple, e.g. $(4,3,5)$.
  3. In 2., adding $k$ to $n$ and multiplying $p$ and $z$ by $2^{k}$ will produce a new solution for any positive integer $k$.
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Another simple solution:

x = 1, p = 2, n even, y = n+3

Then, putting m = n/2:

$(2^n)^x + p^y = 2^n + 2^3.2^n = (1 + 2^3).2^n = 9.2^n = 9.4^m = (3.2^m)^2$

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You have given good explanation. can you generalize the method of solving the equation of the type $2^x$ + $19^y$ = $z^2$ has a solution (3, 0, 3). Also, $8^x$ + $19^y$ = $z^2$ has no solution. But, $8^x$ + $17^y$ = $z^2$ has a solution (2, 1, 9). I think these there are very close to each other in forms. But, finding solutions by generalizing is difficult. Can you clarify...plZ. – richard_simson Mar 20 '12 at 5:54
richard I can't see a very close connection between what you are asking and my answer. You might like to post a new question. – Adam Bailey Mar 20 '12 at 9:58
@richard: Please do not use the answer box for anything but answering. You should probably register, and then edit the question to ask for further clarifications. – Mariano Suárez-Alvarez Mar 21 '12 at 0:06

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