- #1

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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?

- Thread starter al-mahed
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- #1

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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?

- #2

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- #3

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- #4

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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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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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- #7

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

EDIT: Nevermind, I misread the problem.

- #11

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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.

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

- #12

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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 anySuch 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.

- #13

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Isn't m^m m to the mth power not m*mSuch 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.

- #14

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

- #15

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it happens with me all the time! to avoid further mistakes we are talking about [itex]2^n+3=m^m[/itex]I see that, as is frequent, I did not read the original problem correctly.

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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[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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[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

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

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