Solving a Tough Math Problem: Any Ideas?

[al-mahed]

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?

 

[iironiic]

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.

 

[jgm340]

If you are right that "m is of the form 4x+3 for n>1", then the only solution is the trivial solution.

 

[robert Ihnot]

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.

 

[al-mahed]

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

 

[robert Ihnot]

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.

 

[al-mahed]

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

 

[robert Ihnot]

What are the solutions to 2^n+5 = m^2? There is some real misunderstanding here.

 

[al-mahed]

hi robert

as for $$2^n+5=m^2\ ==>\ 4(2^{n-2}+1)+1=m^2=4k(k+1)+1$$, 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 $$2^{n-2}+1$$ 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

 

[TylerH]

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.

 

[robert Ihnot]

hi robert

as for $$2^n+5=m^2\ ==>\ 4(2^{n-2}+1)+1=m^2=4k(k+1)+1$$, 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 $$2^{n-2}+1$$ 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.

 

[al-mahed]

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

 

[ramsey2879]

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

 

[robert Ihnot]

I see that, as is frequent, I did not read the original problem correctly.

 

[al-mahed]

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 $2^n+3=m^m$


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

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

$$\large\ m^{2^{n-1}}\equiv\ 1\mod 2^n$$, but $$\large\ m^m\equiv\ 3\mod 2^n$$, hence

$$\large\ m^{2^{n-1}}\equiv\ m^m-2\mod 2^n$$

 

[al-mahed]

the font is too small, in case you cannot see the exponet properly I was trying to say that (now I modified some things):

$$\large\ (m^2)^{n-1}\equiv\ 1\ mod\ 2^n\ ==>\ m^{2^n}\equiv\ m^2 \ mod\ 2^n\ ==>\ m^{(m^m-3)}\equiv\ m^2\ mod\ 2^n$$

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

I don`t know exactly how to proceed from here, but $$m^m-5$$ cannot be a primitive root, otherwise it would imply (by the theorem of primitive roots) that $$m^m-5 | 2^{n-1}$$, and that`s not true for n> 1 since $$2^n-2=m^m-5$$

 

[al-mahed]

just to point ouy that I made a terrible mistake

$$\large\ m^{2^{n-1}}\equiv\ 1\ mod\ 2^n\ ==>\ m^{2^n}\equiv\ m^2 \ mod\ 2^n$$

is false, the correct is

$$\large\ m^{2^{n-1}}\equiv\ 1\ mod\ 2^n\ ==>\ m^{2^n}\equiv\ 1 \ mod\ 2^n$$

since $$(m^2)^{n-1}\neq\ m^{2^{n-1}]$$

 

[al-mahed]

for instance 2^(2^3) = 2^8 = 256

and (2^2)^3 = 4^3 = 64

 

[robert Ihnot]

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.