Problems & Puzzles:
Puzzles
Puzzle 198. 2^a  3^b = c
A few days ago Eric Brier sent the following
"funny puzzle":
Prove that the equation 2^a
 3^b = 103 has no solutions with a and b integers.
________
Note: Eric claims to have a proof of it
Formal and complete solution for this puzzle came from
Johann Wiesenbauer, Ken Wike
and
Santi
Spadaro, basically using the same approach:
2^a3^b = 103 > 2^a = 103+3^b > 2^a == 1 (mod 3) Then a must be
even: put a=2c.
4^c3^b = 103 > 3^b=4^c103 > 3^b == 3 == 1(mod 4) Then b must be even: put b=2d.
So we have 2^(2c)3^(2d) = 103 > (2^c3^d)(2^c+3^d) = 103. Since 103
is prime this can only be if:
2^c3^d = 1
2^c+3^d = 103
Adding the two equations we get 2^(c+1)=104 which can't be since 104 is
not a power of two. Then 2^a3^b = 103 has no solutions in integers.
Another (theoreticalexhaustive) approach came form
Joseph L. Pe, based on the analysis of the finite & exhaustive 3 digits endings of 2^a & 3^b.
Proof. The key to the proof is in the following two
facts:
Fact A. The sequence of the last three digits of 2^n
is eventually periodic with period = 100, that is, 2, 4, 8, ..., 504, 008,
..., 504, 008, ..., 504, .... where 008, ..., 504 is a repeating list of
100 terms. This sequence can be generated by, say, the Mathematica command
Table[Mod[2^n, 1000], {n, 1, 200}].
Fact B. The sequence of the last three digits of 3^n
is eventually periodic with period = 100, that is, 3, 9, 27, ...,
001, 003, ..., 001, 003, ..., 001, .... where 003, ..., 001 is a repeating
list of 100 terms. This sequence can be generated by, say, the Mathematica
command Table[Mod[3^n, 1000], {n, 1, 200}].
Let X be the last three digits of 2^a and Y be the
last three digits of 3^b. In the subtraction of 3^b from 2^a, Y will be
subtracted from X. This subtraction determines the last three digits of
2^a  3^b, which are required to be 1, 0, 3.
But there are less than 200 choices for X and less
than 200 choices for Y. It is not hard to check exhaustively (say, by a
simple computer program) that no choice of X and Y will make 2^a  3^b end
with the digits 1, 0, 3. Therefore, 2^a  3^b = 103 has no integer
solutions.
For example, X can be 648 and Y can be 427, in which
case X  Y = 221, and so 2^a  3^b ends with the digits 221, and cannot
equal 103. For a case with borrowing in the subtraction, consider X as 016
and Y as 923. Then the "0" in X must borrow from its left neighbor in 2^a,
so that 2^a  3^b ends with the digits 093 (= 1016  923), hence cannot
equal 103.
One more approach ("empirical")
came for Felice Russo:
I don't have a theoretical proof but an empirical one. The differences 2^n3^m are the sequences 2^n3^0, 2^n3^1,
2^n3^2, 2^n3^3............. for n=1,2,3,4,5......
By looking at those sequences (please see the
attached file) we can see very easily that for n>8 and m>5 the difference 2^n3^m is always
greater than 103. So the only values that must be checked are very few: 27. None of
them is equal to 103 and then the equality
2^n3^m=103 is never satisfied.
In particular Johann Wiesenbauer could prove a
more general statement than the asked by Brier:
... (here) it is proved that 2^a+3^b=c has no
integer solutions a,b, whenever c is a prime, but not a Mersenne prime,
and c=7 mod 12.
Let's assume that a and b are integer solutions. Viewing this equation
mod 3, it becomes (1)^a=1 mod 3, which proves that a must be even. In
particular, since a must be clearly positive, 2^a=0 mod 4 and hence
(1)^b=1 mod 4 proving that b is even as well. As a consequence we have
the factorization
c = (2^(a/2)3^(b/2))(2^(a/2)+3^(b/2))
and a fortiori
2^(a/2)3^(b/2)=1
2^(a/2)+3^(b/2)=c
since c is a prime and the first factor is clearly smaller than the
second one. By adding these two equations we see that c+1 is a power of 2
contradicting our assumption that c is not a Mersenne prime.
c=7 mod 12 is according to the Chinese Remainder Theorem equivalent to
the two conditions:
c=1 mod 3 and c=1 mod 4
and this is exactly what I substituted for c when viewing the equation
2^a3^b=c mod 3 and mod 4, respectively.
***
Solutions are still coming!
one more from L. T. Pebody and another very close to the first
given above from J.C. Rosa:
There are no solutions to 2^a3^b=103. Look modulo
39. The powers of 2 are 1,2,4,8,16,32,25,11,22,5,10,20. The powers of 3
are 1,3,9,27. 103 is equivalent to 25. Therefore 3^b+103 must be
equivalent to 26,28,34 or 13 modulo 39. No solutions exist. (Pebody)
103=52^251^2=16*(13^2)9*(17^2). So the equation
2^a3^b=103 becomes: 16*(2^(a4)169)=9*(3^(b2)289) From where
2^(a4)169=0 mod 9 and 3^(b2)289=0 mod 16 consequently a and b are
even.
Let a=2*p ; b=2*q . The equation 2^a3^b=103
becomes: (2^p)^2(3^q)^2=103 , but 103 is prime and consequently 2^p=52
and 3^q=51. Impossible . Therefore the equation 2^a3^b=103 has no
solution (Rosa)
***
