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Okay... Here goes nothing.
2^m = 3^n + 5
Since 2^m > 3^n, it follows that m > n [i.e.]
Let f(x)=2^(x+a)-3^x-5. [using]
f(x)=0 & a=2 => x=1,3
f(x)=0 & a=3 => no +ve integer solutions
f(x)=0 & a=4 => no +ve integer solutions
f(x)=0 & a=5 => no +ve integer solutions
etc. [the curve keeps translating & transforming]
Suitable solutions occur when a=2:
x=1 => n=1 => m=n+2=3
x=3 => n=3 => m=n+2=5
(3,1) & (5,3) are the only positive integer solution pairs.
2^m = 3^n + 5
Since 2^m > 3^n, it follows that m > n [i.e.]
Let f(x)=2^(x+a)-3^x-5. [using]
f(x)=0 & a=2 => x=1,3
f(x)=0 & a=3 => no +ve integer solutions
f(x)=0 & a=4 => no +ve integer solutions
f(x)=0 & a=5 => no +ve integer solutions
etc. [the curve keeps translating & transforming]
Suitable solutions occur when a=2:
x=1 => n=1 => m=n+2=3
x=3 => n=3 => m=n+2=5
(3,1) & (5,3) are the only positive integer solution pairs.
dvs
Okay... Here goes nothing.
2^m = 3^n + 5
Since 2^m > 3^n, it follows that m > n [i.e.]
Let f(x)=2^(x+a)-3^x-5. [using]
f(x)=0 & a=2 => x=1,3
f(x)=0 & a=3 => no +ve integer solutions
f(x)=0 & a=4 => no +ve integer solutions
f(x)=0 & a=5 => no +ve integer solutions
etc. [the curve keeps translating & transforming]
Suitable solutions occur when a=2:
x=1 => n=1 => m=n+2=3
x=3 => n=3 => m=n+2=5
(3,1) & (5,3) are the only positive integer solution pairs.
2^m = 3^n + 5
Since 2^m > 3^n, it follows that m > n [i.e.]
Let f(x)=2^(x+a)-3^x-5. [using]
f(x)=0 & a=2 => x=1,3
f(x)=0 & a=3 => no +ve integer solutions
f(x)=0 & a=4 => no +ve integer solutions
f(x)=0 & a=5 => no +ve integer solutions
etc. [the curve keeps translating & transforming]
Suitable solutions occur when a=2:
x=1 => n=1 => m=n+2=3
x=3 => n=3 => m=n+2=5
(3,1) & (5,3) are the only positive integer solution pairs.
This isn't really a proof though, as there should be solutions (and positive ones) for each value of a and you haven't really proved that they're not integers. Having said that, I can't figure it out either (but I have discovered things like both m and n must be odd due to modulo arithmetic equations, but didn't get much further than that towards a full proof).
If m is even then LHS = 1 mod 3.
But RHS = 0 mod 3 + 2 mod 3 = 2 mod 3 so m is odd.
If n is even (and m is odd), then set m = 2a+1, n = 2b.
LHS = 2(4^a) = 0 mod 4
RHS = 9^b + 5 = 1 mod 4 + 1 mod 4 = 2 mod 4 so n is odd.
We see also that m>n
This is all I can do.
But RHS = 0 mod 3 + 2 mod 3 = 2 mod 3 so m is odd.
If n is even (and m is odd), then set m = 2a+1, n = 2b.
LHS = 2(4^a) = 0 mod 4
RHS = 9^b + 5 = 1 mod 4 + 1 mod 4 = 2 mod 4 so n is odd.
We see also that m>n
This is all I can do.
RichE
It was a conjecture of Pillai, now a theorem of Stroeker and Tijdeman from 1982, that
|3^m - 2^n| = c
has at most one solution for c > 13.
The exceptions are c=1,5,7,13,23 and when c=5 the only solutions are
(m,n) = (2,2), (1,3), (3,5).
|3^m - 2^n| = c
has at most one solution for c > 13.
The exceptions are c=1,5,7,13,23 and when c=5 the only solutions are
(m,n) = (2,2), (1,3), (3,5).
If you mean these to be the solution to the original problem, we can see that since m>n (or simply by substituting) , none of these pairs work.
RichE
Because of the modulus sign in the equation, the result I gave deals with all solutions to the equations
3^m - 2^n = 5
and
3^m - 2^n = -5.
My point was only that this was, until relatively recently, an open problem and is near the cutting edge of research in number theory.
3^m - 2^n = 5
and
3^m - 2^n = -5.
My point was only that this was, until relatively recently, an open problem and is near the cutting edge of research in number theory.
2^2<>3^2+5 so not a solution of 2^m = 3^n + 5
heres my effort:
2^m = 3^n + 5
only has solutions (5,3) (3,1)
2^4=3^4 mod 5
so 2^4m= 3^4m mod 5
2^4m+1=2 mod 5 3^4m+1=3 mod 5
2^4m+2=4 mod 5 3^4m+2=4 mod 5
2^4m+3=3 mod 5 3^4m+3=2 mod 5
so possible solutions are of the form
m=4k n=4l (1)
m=4k+1 n=4l+3 (2)
m=4k+2 n=4l+2 (3)
m=4k+3 n=4l+1 (4)
for case (1) (and sim argument for case (3) ) would have
2^4k=3^4l=5
2^4k-3^4l=5=(2^2k+3^2l)(2^2k-3^2l) a contradiction since 5 is prime
for case (2)
have 2^(4k+1)=3^(4m+3)+5
so 3^(4m+3)+5 has prime decomposition =2^(4k+1)
but consider 3^(4m+3)+5 mod 2^5=32
3^(4m+3)+5=3^3.(3^4)^m+5
=27.(81)^m+5=27.(17)^m+5 mod 32
=5(1-(-15)^m) mod 32 (27=-5 17=-15 mod32)
this has a last digit of 0 for all m>0
but mod 16 3^(4m+3)+5=11.(1)^4+5=0
so either 3^(4m+3)+5 does not equal 0 mod 32 or it contains prime factors other than 2 as we can divide by 16 which leaves us an odd number.contradiction.
so 3^(4m+3)+5 not equal 0 mod 32
which implies
4k+1<5 and so k<1 which gives k=0
which does not give a solution (it gives m=2 which relates to a solution of previous post) so only solution is
for m=0 3^3+5=32=0 mod 32
so m=0 gives a solution which relates to (5,3)
for case (4)
have 2^(4k+3)=3^(4m+1)+5
consider 3^(4m+1)+5 mod 2^4=16
=3.1^4m+5
=8 for all m
which implies 4k+3<4 so k<1/4 but k integral so k=0 only solution
this relates to solution (3,1)
Can anyone who looks at this and agrees with it find my post on mind maps and investigations and please post some replies!!
)for case (2)
have 2^(4k+1)=3^(4m+3)+5
so 3^(4m+3)+5 has prime decomposition =2^(4k+1)
but consider 3^(4m+3)+5 mod 2^5=32
3^(4m+3)+5=3^3.(3^4)^m+5
=27.(81)^m+5=27.(17)^m+5 mod 32
=5(1-(-15)^m) mod 32 (27=-5 17=-15 mod32)
this has a last digit of 0 for all m>0
but mod 16 3^(4m+3)+5=11.(1)^4+5=0
so either 3^(4m+3)+5 does not equal 0 mod 32 or it contains prime factors other than 2 as we can divide by 16 which leaves us an odd number.contradiction.
so 3^(4m+3)+5 not equal 0 mod 32
Erm, i don't quite understand this part, i'm probably just missing something but 3^11 + 5 = 177152 = 0 mod 2^10
edit: i know that this still has 173 as a factor, but i don't get how your proof has shown that 3^(4m+3) + 5 neccessarily has an odd factor if it's 0 mod 32.
I wish there was a smiley that showed a head with something going *whoosh* over it. That would sum up my feelings about this thread.
I am full of repect for Maths people - I just don't understand it at all!
I am full of repect for Maths people - I just don't understand it at all!
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