# Solutions of $x^2 + 119 = 15 \cdot 2^n$ without trial and error

I seen this equation at math.stack exchange

The equation $x^2 + 119 = 15 \cdot 2^n$ has only six solutions. Those are $(1,3) ,(11, 4), (19, 5), (29, 6), (61, 8)$ and other one is I don't know. This question I have seen in this site. There they given that, it has six solutions and they did not list these all solutions. I got $5$ of the solutions by my trail method. But, may be there is one more solution. Now my question is, how to find these solutions without using computer or calculator to determine these solutions and how one can conclude the number of solutions are six or some n?

Thank you for providing the last solution (-1, 3). But, again by computation in trail and error I got the sixth solution. (701, 15). Now, I understand that, the above equation has 6 solutions in positive integers. So, we can consider this equation as Diophantine equation. Now my question, how to generalize this equations for finding the solutions?

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no one is responding to my question. If any one can, plz answer for generalizing the solution of my post without trail and error method. – gandhi Mar 24 '12 at 10:32
Did you read the paper that I suggested? – Kirthi Raman Mar 24 '12 at 13:39

Gandhi, there is a proof and therefore I am not going to reproduce that and claim it is mine. You may read it at here from the end of page 1 follow page 4. It shows there are only $6$ solutions.

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RAMAN! Extraordinary help you given. Thank you so much. can you provide your mail id sir. plz... – gandhi Mar 24 '12 at 14:50
@RAMAN! Sir, let p > = 7 be a prime number. Find the triples (x, y, z) in Z such as xyz not equal to zero and gcd (x, y, z) = 1 and $x^p$ + 2$y^p$ = $z^2$. I want triplets and proof/generalization. The reason for asking here, I am in position to construct equations and finding solutions by trail method. I am not in position to construct a proof or good generalizations. I hope, with your help, I can end. – gandhi Mar 24 '12 at 14:56
@ALL MEMBERS! please do not misunderstanding me for especially using the name RAMAN SIR. The reason for writing RAMAN Sir, before me new question is, he found an excellent proof. I am interested in learning the proof. If any one can help me in this aspect. I am so thankful to them. – gandhi Mar 24 '12 at 14:59
@Gandhi on math.se there are many many learned people. The best way is to post any question here. I am just a drop here. (But I am flattered) – Kirthi Raman Mar 24 '12 at 17:05
@RAMAN thank you so much. – gandhi Mar 24 '12 at 17:52

look because $x^2$ then it could be also -1,so last solution is (-1,3)

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!thank you for your solution. I got sixth solution in positive integers. Please see my new edited part. – gandhi Mar 24 '12 at 8:11
ok good lucks,it was interesting – dato datuashvili Mar 24 '12 at 8:34
@dato doesn't that mean not only $(-1,3)$, but if negative values are allowed then every pair can have a corresponding pair $(x,n)$ with negative values. I think he is looking for only positive values. – Kirthi Raman Mar 24 '12 at 12:17