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2000 JBMO Problems/Problem 2

Revision as of 23:12, 3 December 2018 by KRIS17 (talk | contribs) (Created page with "== Solution == After rearranging we get: <math>(k-n)(k+n) = 3^n</math> Let <math>k-n = 3^a, k+n = 3^{n-a}</math> we get: <math>2n = 3^a(3^{n-2a} - 1)</math> or, <math>(2n/(...")
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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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