2022 IMO Problems/Problem 5
Problem
Find all triples of positive integers with
prime and
Video solution
https://www.youtube.com/watch?v=-AII0ldyDww [Video contains solutions to all day 2 problems]
Solution
Case 1:
- Since
is indivisible by
, then
must also be indivisible by
.
- If
, then
is divisible by
, so
must be a divisor of
, but
obviously has no solutions and we ruled out
already. For
, let's show that there are no solutions using simple inequalities.
- If
, then
and
by throwing away the remaining (non-negative) terms of binomial theorem. For any solution,
, which is impossible for
. That leaves us with
and
, but
for any integer
(proof by induction), so there are no solutions.
- If
, RHS is at most
and LHS is at least
(again from binomial theorem), which gives no solutions as well.
Case 2:
- Since
is divisible by
, then
must also be divisible by
.
- In addition, RHS is at most
, so
. We may write
, where
.
- Since
is a divisor of
and
it must also be a divisor of
, so
and
. We're looking for solutions of
.
- Let's factorise
: if
with
odd and
, it's
- Since
and
contain an odd number of odd terms (remember the assumption
aka
), they're odd. Also,
modulo
, so
and each following term is even but indivisible by
. The highest power of
dividing
is therefore
where
is the highest power dividing
.
- In comparison,
has factors
,
etc (up to
), and
other even factors, so it's divisible at least by
. Since
for
, the only possible solutions have
or
.
- If
, we reuse the inequalities
and
to show that there are no solutions for
.
- Finally,
isn't a factorial,
and
.
Case 3:
Just like in case 2, is divisible by
so
must also be divisible by
. However,
and
are also both divisible by
, so remainders modulo
tell us that no solutions exist.
Conclusion:
The only solutions are .