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Find all pairs of positive integers $(a,b)$ that satisfy $13^a+3=b^2$.

If $a$ is even then $3=(b-13^{a/2})(b+13^{a/2})$ which have no solutions. Now if the case $a=2k+1$ is odd then $b^2-13.(13^k)^{2}=3$ I cant proceed from here, please help. Any other methods for the latter case?

  • a01 and b=4 are solutions – Jose Garcia May 09 '17 at 18:15
  • Well they are obvious but I want a proof weather they are the only solutions or not. –  May 09 '17 at 18:16
  • 13^a + 3 should be a perfect square for satisfying the condition.Does this approach leads to anything? – Savitar May 09 '17 at 18:19
  • no reason to expect any easy proof, although elliptic curves may not be necessary: https://en.wikipedia.org/wiki/Ramanujan%E2%80%93Nagell_equation WHERE did you get this problem? – Will Jagy May 09 '17 at 18:23
  • Someone asked me said it was an easy Olympiad number theory. –  May 09 '17 at 18:36
  • You can solve a similar equation $5^n=6m^2+1$ in integers using Pell equations. See http://vjimc.osu.cz/history 2015 Category II Solutions and http://mathworld.wolfram.com/PellEquation.html. I have an answer here that explains the solutions of one Pell equation. – user236182 Jul 24 '17 at 14:22

1 Answers1

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A solution using elliptic curves, or Mordell equations, as Will Jagy suggests in the comments.

There are three cases:

$1)$ $\ $ $a=3k$, $k\in\mathbb Z^+$. Then $\left(13^k\right)^3+3=b^2$ is a Mordell equation.

See http://oeis.org/search?q=mordell&language=english&go=Search,

in particular (this link is inside the link), e.g., http://oeis.org/A001014/a001014.txt

$b=\pm 2$, $13^k=1$, $k=0$, $a=0$. No solutions, because $k\in\mathbb Z^+$.

$2)$ $\ $ $a=3k+1$, $k\in\mathbb Z$, $k\ge 0$.

Then $\left(13^{k+1}\right)^3+507=(13b)^2$.

$13^{k+1}=13$, $13b=\pm 52$, $k=0$, $a=1$, $b=\pm 4$.

$3)$ $\ $ $a=3k+2$, $k\in\mathbb Z$, $k\ge 0$.

Then $3\cdot 13^4>10000$, so a different method is needed.

Then $13^3\equiv 1\pmod{61}$, $13^2\equiv 47\pmod{61}$

$\left(13^{3}\right)^k\cdot 13^2+3\equiv b^2\pmod{61}$

$50\equiv b^2\pmod{61}$

$2\equiv \left(b\cdot 5^{-1}\right)^2\pmod{61}$

because $\gcd(5,61)=1$.

A contradiction, because $2$ is not a quadratic residue mod $61$, because $61$ is a prime of the form $8k+5$. See Quadratic Reciprocity.

user236182
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  • My friend betrayed me!!!! –  May 09 '17 at 18:50
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    @DarthVader1056 it is possible your friend believed what he said, and the fault lies a few people over. It really is best if each question posted has a very specific source, something in print. People are so fond of contest problems, they post almost anything here, and we rarely know the source of the problem or what background the student has that relates to the problem. – Will Jagy May 09 '17 at 18:53
  • I completely agree with you, once you know the level of contest its easy to think in what tools should one deal with. I am really sorry I dont have any known source for this one. I will make sure to include sources whenever I post in future. –  May 09 '17 at 18:55