Solve 2^n+3=m^m

  • Thread starter al-mahed
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  • #1
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I came across this problem on another forum. Seems to be very tough.

The only trivial facts are:

m is of the form 4x+3 for n>1

a trivial solution (n,m) is (0,2)

for n even, m^m-1 is divisible by 3
for n odd, m^m-2 is divisible by 3


any ideas?
 

Answers and Replies

  • #2
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To input something in this discussion, aside from the trivial solution, m can't be even. If m was even, the parities don't match up, so you know that m has to be odd. I'll try to figure something more later.
 
  • #3
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If you are right that "m is of the form 4x+3 for n>1", then the only solution is the trivial solution.
 
  • #4
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M can not be even, since 2^n+3 is always odd. but all odd squares are congruent to 1 Mod 4, so that2^n congruent to 0 for n greater than one. So he only possibility is m^2 = 5, which is not a square.

Now if we try the case of 2^n+5 =m^2, we have a solution for n=2, m=3. Are there more? No, why?

M can not be even, so it is odd. All such square are congruent to 1 Mod 8. So we have
2^n+5=1+8K, or 2^n+4 = 8K, which reduces to 2^(n-2) + 1 =2K. This is possible only for the previously described case n-2 = 0.
 
  • #5
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robert

I didn`t understand what you did, m is not necessarely of the form 8t+1, m^m is not a square as the exponent m is odd, 3^3 - 1 is not divisible by 8 (and not by 4 neither), 5^5-1 is not divisible by 8, and infinitely many other examples.

jgm340

when n > 1, m is of the form 4x+3

2^1+3=5=4+1
2^2+3=7=4+3
2^3+3=11=4*2+3
2^4+3=19=4*4+3

and so on
 
  • #6
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All odd squares are of the form 8K+1. (2n+1)^2 =4n^2+4n+1 =4n(n+1)+1. I am also looking at the case not for 3, but for 5: 2^n+5 = m^2.
 
  • #7
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I know this fact about perfect odd squares, the thing is: m^m is NOT a perfect odd square, m is odd, so we have (2k+1)^{(2k+1)}=(2k+1)^{2k}(2k+1), a square times a non square (only if m is a square itself)

and moreover, 2^n+3=2^n+2+1=2(2^{n-1}+1)+1, it cannot have the form 4r+1

ps: why are you trying to solve 2^n+5=m^2? it is not the original problem! but if you are really interested on it, you`ll find infinitely many solutions
 
  • #8
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What are the solutions to 2^n+5 = m^2? There is some real misunderstanding here.
 
  • #9
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hi robert

as for [tex]2^n+5=m^2\ ==>\ 4(2^{n-2}+1)+1=m^2=4k(k+1)+1[/tex], in fact there is not infinitely many solutions as I prematurely said, n = 2 is the only solution, as k(k+1) is always even and [tex]2^{n-2}+1[/tex] is even only for n = 2

going back to the original problem, I don`t understand your solution since m^m is never a square, could you please elaborate it more?

cheers
 
  • #10
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What are the solutions to 2^n+5 = m^2?
m=4 and n=3 is the only solution.

EDIT: Nevermind, I misread the problem.
 
  • #11
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hi robert

as for [tex]2^n+5=m^2\ ==>\ 4(2^{n-2}+1)+1=m^2=4k(k+1)+1[/tex], in fact there is not infinitely many solutions as I prematurely said, n = 2 is the only solution, as k(k+1) is always even and [tex]2^{n-2}+1[/tex] is even only for n = 2

going back to the original problem, I don`t understand your solution since m^m is never a square, could you please elaborate it more?

cheers
Such problems are normally worked by making the assumption there is a solution, i.e. that m*m is a perfect squre, that is an integral square.
 
  • #12
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Such problems are normally worked by making the assumption there is a solution, i.e. that m*m is a perfect squre, that is an integral square.
Ok, let`s put into these terms: considering a natural n >0, and (hence!) an odd integer m, find solutions to 2^n+3=m^m, or prove there is any
 
  • #13
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Such problems are normally worked by making the assumption there is a solution, i.e. that m*m is a perfect squre, that is an integral square.
Isn't m^m m to the mth power not m*m
 
  • #14
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I see that, as is frequent, I did not read the original problem correctly.
 
  • #15
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I see that, as is frequent, I did not read the original problem correctly.
it happens with me all the time! to avoid further mistakes we are talking about [itex]2^n+3=m^m[/itex]


I think we can work it out using [itex]a^{\varphi (m)}\equiv\ 1\mod m[/itex]

as m is not divisible by 2, and [itex]\varphi (2^n)=2^{n-1}[/itex] we have

[tex]\large\ m^{2^{n-1}}\equiv\ 1\mod 2^n[/tex], but [tex]\large\ m^m\equiv\ 3\mod 2^n[/tex], hence

[tex]\large\ m^{2^{n-1}}\equiv\ m^m-2\mod 2^n[/tex]
 
Last edited:
  • #16
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the font is too small, in case you cannot see the exponet properly I was trying to say that (now I modified some things):

[tex]\large\ (m^2)^{n-1}\equiv\ 1\ mod\ 2^n\ ==>\ m^{2^n}\equiv\ m^2 \ mod\ 2^n\ ==>\ m^{(m^m-3)}\equiv\ m^2\ mod\ 2^n[/tex]

then [tex]m^2[m^{(m^m-5)}-1]\equiv\ 0\ mod\ 2^n\ ==>\ m^{(m^m-5)}\equiv\ 1\ mod\ 2^n[/tex]

I don`t know exactly how to proceed from here, but [tex]m^m-5[/tex] cannot be a primitive root, otherwise it would imply (by the theorem of primitive roots) that [tex]m^m-5 | 2^{n-1}[/tex], and that`s not true for n> 1 since [tex]2^n-2=m^m-5[/tex]
 
  • #17
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just to point ouy that I made a terrible mistake

[tex]\large\ m^{2^{n-1}}\equiv\ 1\ mod\ 2^n\ ==>\ m^{2^n}\equiv\ m^2 \ mod\ 2^n[/tex]

is false, the correct is

[tex]\large\ m^{2^{n-1}}\equiv\ 1\ mod\ 2^n\ ==>\ m^{2^n}\equiv\ 1 \ mod\ 2^n[/tex]

since [tex](m^2)^{n-1}\neq\ m^{2^{n-1}][/tex]
 
  • #18
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for instance 2^(2^3) = 2^8 = 256

and (2^2)^3 = 4^3 = 64
 
  • #19
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2^n+3 = m^m, then 2^n +4 = m^m+1 = (m+1)(1-m+m^2-+++m^(m-1)). We know m is odd, and since the later part of the factorization is odd, we have 4 and only 4 divides m+1, so m==3 Mod 8.
 

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