Diophantine Equations(1)
by XmL, Apr 3, 2014, 4:01 AM
These blog posts are just for personal documentation of problems, so feel free to add your own solutions or point out my mistakes.
Problem: Find all positive integers
such that
.
Solution
Problem: Find all positive integers


Solution
We first rearrange so that the same variables are on the same side:
. Notice that
is a perfect square and if so is
, then we can factor the LHS, this motivates me to discuss the parities of
.
If
is odd, then when
,
. When
, if
, then
and
which is impossible since
is odd. If
, there's no solution. Hence
is the only solution when
is odd.
If
is even, let
, where
. Our equation is now:
. We now take the usual analysis: Let
. From (1)+(2):
, since
is not divisible on the LHS, it must be true that
(if
, then (2)>(1) which is impossible). Hence
.
Now the value of
depends on the largest power of
on the RHS when
. On the other hand, (2)-(1) gives
(this implies
,
). If
or
, we can deduce that there's no solution: ${2^{a-3}3^n=a-1+2^{n-a-3}$ or
, note that
.
Now it just leaves the case when
or
. This means
. Since
is divisible by
but not any larger powers, therefore
is the largest power that divides the RHS, therefore
where
is the only positive solution
, this solution is shown above that it leads to no solution. Hence in conclusion
is our only solution.




If











If










Now the value of










Now it just leaves the case when










This post has been edited 3 times. Last edited by XmL, Aug 12, 2014, 11:19 PM