solve in positive integers: 3 \cdot 2^x +4 =n^2

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solve in positive integers: 3 \cdot 2^x +4 =n^2

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Source: Greece JBMO TST 2019 p2

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#1 h Apr 29, 2019, 12:39 AM • 3 Y

Y by Iora, ItsBesi, Adventure10

Find all pairs of positive integers $(x,n)$ that are solutions of the equation $3 \cdot 2^x +4 =n^2$ .

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#2 h Apr 29, 2019, 1:49 AM • 3 Y

Y by A-Thought-Of-God, Iora, Adventure10

Solution

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#3 h Nov 8, 2022, 7:29 PM • 1 Y

Y by hanzo.ei

Write the given expression as $2^x \cdot 3=(n+2)(n-2)$ . Taking $\bmod 3$ on both sides we get $n \equiv 1,3 (\bmod 3)$ .
$\underline{\textbf {Case 1:}}$ If $n=3k+1$ , then replace $n$ to get $2^x=(k+1)(3k-1)$ . If $k$ is odd then $k+1$ and $3k-1$ both are even. For integer $k$ , $k+1 \leq 3 k-1$ and as they are perfect powers of $2$ , so $k+1|3 k-1 \Rightarrow k+1| 3 k-1-3 k-3 \Rightarrow k+1 \mid 4$ $\implies k+1=\pm 4, \pm 2, \pm 1$ . Of these, for $k+1= 2,4$ gives $n=4,10$ . Plugging these into the original equation gives $\boxed{(x,n)=(2,4), (5,10)}$ .
If $k$ is even then $k+1$ odd, so $k+1= 2^0=1 \implies n=1$ which yields no solution.
$\underline{\textbf {Case 2:}}$ $n=3k+2$ . Plugging in the value of $n$ , we get $k(3k+4)=2^x$ . Again, by similar logic as before $k \mid 3k+4 \implies k \mid 4$ . So, $k=1,2,4$ . From these, for $k=4$ we get $n=14$ and plugging in the value, we get $\boxed{(x,n)=(6,14)}$

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#4 h Wednesday at 7:39 PM

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Trivial

This post has been edited 4 times. Last edited by ali123456, Wednesday at 7:49 PM

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