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Difference between revisions of "2000 JBMO Problems/Problem 2"

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Thus <math>n = 1</math> or <math>3</math>.
 
Thus <math>n = 1</math> or <math>3</math>.
  
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~Kris17
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<math>Kris17</math>
 

Revision as of 22:48, 12 March 2023

Problem 2

Find all positive integers $n\geq 1$ such that $n^2+3^n$ is the square of an integer.


Solution

After rearranging we get: $(k-n)(k+n) = 3^n$

Let $k-n = 3^a, k+n = 3^{n-a}$

we get: $2n = 3^a(3^{n-2a} - 1)$ or, $(2n/(3^a)) + 1 = 3^{n-2a}$

Now, it is clear from above that $3^a$ divides $n$. so, $n \geq 3^a$


If $n = 3^a, n - 2a = 3^a - 2a \geq 1$ so $RHS \geq 3$ But $LHS = 3$

If $n > 3^a$ then $RHS$ increases exponentially compared to $LHS$ so $n$ cannot be $> 3^a$.

Thus $n = 3^a$.

Substituting value of $n$ above we get:

$3 = 3^{3^a - 2a}$

or $3^a - 2a = 1$ this results in only $a = 0$ or $a = 1$

Thus $n = 1$ or $3$.

~Kris17

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