Difference between revisions of "2015 IMO Problems/Problem 2"
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<math>2^n = ac-b \geq (a-1)c \geq 2</math>, i.e., <math>n</math> and <math>p</math> are positive. | <math>2^n = ac-b \geq (a-1)c \geq 2</math>, i.e., <math>n</math> and <math>p</math> are positive. | ||
− | + | Observe that if <math>a=b\geq 3</math>, we get <math>a(c-1)=2^n</math>, so <math>a</math> and <math>c-1</math> are (even and) | |
powers of <math>2</math>. Hence <math>c</math> is odd and <math>a^2-c=2^m=1</math>. Hence <math>c+1=a^2</math> is also a | powers of <math>2</math>. Hence <math>c</math> is odd and <math>a^2-c=2^m=1</math>. Hence <math>c+1=a^2</math> is also a | ||
power of <math>2</math>, which implies <math>c=3</math>. But <math>a=b=c=3</math> is not a | power of <math>2</math>, which implies <math>c=3</math>. But <math>a=b=c=3</math> is not a |
Latest revision as of 22:51, 16 October 2015
Problem
Determine all triples of positive integers such that each of the numbers
is a power of 2.
(A power of 2 is an integer of the form where
is a non-negative integer ).
Solution
The solutions for are
,
,
,
, and
permutations of these triples.
Throughout the proof, we assume ,
so that
,
,
, with
. Note that
since
otherwise
, which is impossible. Hence
, i.e.,
and
are positive.
Observe that if , we get
, so
and
are (even and)
powers of
. Hence
is odd and
. Hence
is also a
power of
, which implies
. But
is not a
solution; hence
is infeasible. We consider the remaining cases as follows.
Case 1: . We have
From the second equation,
is even. From the third equation, if
, then
; if
, then
is odd, which implies that
. Hence
(so
),
, and
. Hence
. Hence
is 2 or 4, and
equals
or
. Thus the solutions for
are
,
or
.
Case 2: .
Since
, we have
. Hence
Hence
is not divisible by
, and
is not divisible
by
for
. Adding and subtracting
and
, we get
From the latter equation,
is divisible by
. Hence
is not
divisible by
, which implies that
is a multiple of
.
Hence
and
.
Consider , which implies
,
,
.
Hence
, or
. Hence
,
,
and
.
Finally, consider ,
,
. Hence
. But
implies
and
implies
. Hence there are no
solutions with
.
We obtain as the only solution with
.
See Also
2015 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |