Difference between revisions of "2000 JBMO Problems/Problem 2"
(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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+ | ==Problem 2== | ||
+ | |||
+ | Find all positive integers <math>n\geq 1</math> such that <math>n^2+3^n</math> is the square of an integer. | ||
+ | |||
+ | |||
== Solution == | == Solution == | ||
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Thus <math>n = 1</math> or <math>3</math>. | Thus <math>n = 1</math> or <math>3</math>. | ||
+ | ~Kris17 | ||
+ | |||
+ | == Solution 2 (credit to dskull16) == | ||
− | <math> | + | n = 1 is an obvious solution but are there any more? |
+ | We require that <math>n^2 + 3^n = (n+k)^2</math> for some k in the naturals. Using difference of two squares and realising that the factor pairs can only be a power of 3, we get that | ||
+ | <math>2n+k = 3^{n-j}, k = 3^j</math> which gives us <math>2n = 3^{n-j} - 3^j</math>. While we could consider induction on j to prove that <math>3^{n-j} - 3^j > 2n</math>, we could instead consider the difference between <math>3^n</math> and all the powers of 3 preceding it. The smallest difference between the nth power of 3 and any other power of 3 before it is trivially the n-1th power of 3 so it suffices to show that: | ||
+ | <math>3^n - 3^{n-1} > 2n</math> for <math>n > 1</math>, which simplifies to <math>2\cdot 3^{n-1} > 2n</math> and hence <math>3^{n-1} > n</math> which is trivially true <math>\forall n > 1</math>. Hence there are no further solutions. |
Latest revision as of 20:24, 21 February 2024
Problem 2
Find all positive integers such that
is the square of an integer.
Solution
After rearranging we get:
Let
we get:
or,
Now, it is clear from above that divides
. so,
If
so
But
If then
increases exponentially compared to
so
cannot be
.
Thus .
Substituting value of above we get:
or this results in only
or
Thus or
.
~Kris17
Solution 2 (credit to dskull16)
n = 1 is an obvious solution but are there any more?
We require that for some k in the naturals. Using difference of two squares and realising that the factor pairs can only be a power of 3, we get that
which gives us
. While we could consider induction on j to prove that
, we could instead consider the difference between
and all the powers of 3 preceding it. The smallest difference between the nth power of 3 and any other power of 3 before it is trivially the n-1th power of 3 so it suffices to show that:
for
, which simplifies to
and hence
which is trivially true
. Hence there are no further solutions.