Turkey MO #4
by XmL, Jun 5, 2014, 3:43 AM
Problem: Find all positive integers
and
satisfying 
Solution
I did this problem in 30 minutes after a really boring math exam.



Solution
The moment I saw
I was motivated to approach this problem by analyzing the powers of two. In this solution, define
the maximum power of
that divides
.
Since
, hence
because if not, let
, then divide both sides by $min({2^{q(m!)},2^{n})$ to get a contradiction, the same thing can be done for
. Note that this condition means
.(This suggests a finite number of solution and we can prove that using inequality).
It's convenient to make a list of values for
. Testing starting with small values gives
. At
it's pretty clear that there's no solution for larger values.
We claim that
for
.
To prove this we have to find a lower bound for
in terms of
so that the inequality is more manageable. One well-known way to calculate
is
.
We can check by hand that our claim is true for
.
For
.
We now proceed by proving
. We take
on both sides:
. Since
(note that
), we increase our bound:
which is true for our domain.
Hence $2^{2^{q(m!)}}+2^{q(m!)}\ge 2^{2^{\frac {m}{2}+2}}+2^{\frac {m}{2}+2}}>m!$ for all
, and this means our claim has been proven.
In conclusion
is the only solution.




Since





It's convenient to make a list of values for



We claim that


To prove this we have to find a lower bound for




We can check by hand that our claim is true for

For

We now proceed by proving






Hence $2^{2^{q(m!)}}+2^{q(m!)}\ge 2^{2^{\frac {m}{2}+2}}+2^{\frac {m}{2}+2}}>m!$ for all

In conclusion

I did this problem in 30 minutes after a really boring math exam.
This post has been edited 1 time. Last edited by XmL, Aug 12, 2014, 11:17 PM