[QUOTE=R.D. Silverman;257341]Huh???. m = 1,2,3,4,5, or 6 are not 0 mod 7, yet m < 7. Surely, this
can't be what you meant?

m!= 0 mod 7 is a necessary condition for 24m+7 to be prime except for the
special (degenerate case) m = 0. Otherwise 24m+7 will be divisible by 7.
I believe that m > 0 was part of the conditions.




Complete idiot. 7 certainly does equal 0 mod 7.



You have ZERO idea of what a modular function is. Try google. I will
give some hints:

(1) It is a meromorphic function in the upper half plane.
(yes, I know. You are lost already)
(2) The function involves the modular group; an instance of
SL(2,Z) (still lost......)
(3) Do you know what a linear fractional transformation is?[/QUOTE]

Here's an exercize for you.

Prove (or disprove) that a linear fractional transform under the modular
group is a conformal map of the upper half plane.

[QUOTE=R.D. Silverman;257343]Here's an exercize for you.

Prove (or disprove) that a linear fractional transform under the modular
group is a conformal map of the upper half plane.[/QUOTE]

and easier way would be that z=x mod y shows that z|y with remainder x. and then go on to show 7|7 with remainder 0.

[QUOTE=R.D. Silverman;257341]Huh???. m = 1,2,3,4,5, or 6 are not 0 mod 7, yet m < 7. Surely, this
can't be what you meant?

m!= 0 mod 7 is a necessary condition for 24m+7 to be prime except for the
special (degenerate case) m = 0. Otherwise 24m+7 will be divisible by 7.
I believe that m > 0 was part of the conditions.


Complete idiot. 7 certainly does equal 0 mod 7.

[/QUOTE]

Ah, yet you still miss what I was trying to say. The statement I was making to clarify other possibilities beyond m!=0 mod 7. When m=1, m!=1 mod 7, and 24m+7 IS prime. When m=3, m!=6 mod 7, and 24m+7 IS prime. What I was attempting to point out is that m! does not, but may, equal 0 mod 7 for 24m+7 to be prime. Also, m!=0 mod 7 for all m's greater than or equal to 7. You will note that 8!=40320, which is 0 mod 7, as is 9!, 10!, 11!, et al. Yes, et al. Taking n>x, n!=0 mod x, where both are whole numbers.

The "necessary condition" is that m is not a multiple of 7. If m IS a multiple of 7 (we'll use 77 as a generic example), then 24m+7 is evenly divisible by 7.
See first year college algebra for the logic used. 24*77+7=1855, which is 5*[B]7[/B]*53.

This is not a requirement, but merely a poor musing that put no thought into the question. 24(1)+7=31
24(3)+7=79
24(4)+7=103
24(5)+7=127
24(6)+7=151
((24(7)+7=175)) though 7!=0 mod 7

when m=1,3,4,5, or 6, m! does not equal 0 mod 7, yet the result is prime. However, when 7 is used for m, m!=0 mod 7 as you earlier stated as a requirement, but does not make the equation prime.

[QUOTE=c10ck3r;257386]This is not a requirement, but merely a poor musing that put no thought into the question. 24(1)+7=31
24(3)+7=79
24(4)+7=103
24(5)+7=127
24(6)+7=151
((24(7)+7=175)) though 7!=0 mod 7

when m=1,3,4,5, or 6, m! does not equal 0 mod 7, yet the result is prime. However, when 7 is used for m, m!=0 mod 7 as you earlier stated as a requirement, but does not make the equation prime.[/QUOTE]

!= means "not equal" in programming not "factorial equal". A easy way to remember this is "! =" is "factorial equals" and "!=" is "not equal""

[QUOTE=c10ck3r;257385]Ah, yet you still miss what I was trying to say. The statement I was making to clarify other possibilities beyond m!=0 mod 7. When m=1, m!=1 mod 7, and 24m+7 IS prime.

[/QUOTE]

Duh! m = 1 mod 7 means that m!= 0 mod 7.

Look, moron. There are only two possibilities. Either m = 0 mod 7,
in which case 24m+7 is prime only for m = 0, or m != 0 mod 7, in which
case 24m + 7 can be prime.

[QUOTE]
When m=3, m!=6 mod 7, and 24m+7 IS prime. What I was attempting to point out is that m! does not, but may, equal 0 mod 7 for 24m+7 to be prime.
[/QUOTE]

Digging yourself deeper and deeper. if m = 0 mod 7, then m = 7K
whence 24(7K) + 7 is divisible by 7 and hence NOT prime except for
7 itself (m = 0)

<plonk>

You are a total cretin and don't even realize it.

[QUOTE]
Also, m!=0 mod 7 for all m's greater than or equal to 7.
[/QUOTE]

Another idiotic statement. m = 14 is certainly greater than 7 and is
certainly 0 mod 7.

[QUOTE=R.D. Silverman;257401]Duh! m = 1 mod 7 means that m!= 0 mod 7.

Look, moron. There are only two possibilities. Either m = 0 mod 7,
in which case 24m+7 is prime only for m = 0, or m != 0 mod 7, in which
case 24m + 7 can be prime.



Digging yourself deeper and deeper. if m = 0 mod 7, then m = 7K
whence 24(7K) + 7 is divisible by 7 and hence NOT prime except for
7 itself (m = 0)

<plonk>

You are a total cretin and don't even realize it.



Another idiotic statement. m = 14 is certainly greater than 7 and is
certainly 0 mod 7.[/QUOTE]

I think what's going on is that they are mistaking m is not equal to 0 mod 7 for m factorial is equal to 0 mod 7. I get this from:

[QUOTE=c10ck3r;257385] 8!=40320[/QUOTE]

8 factorial is equal to 40320 so they are thinking the ! is factorial when it's the programming for not.

Bob-I apologize for the misunderstanding. However, on a MATH forum, one would expect symbols to take their math definitions and not code form. Next time, please clarify by using mathematics in math forums, even if only in brackets. In case you were unaware, it is unicode 2260. Sorry for the confusion.

[QUOTE=c10ck3r;258322]Bob-I apologize for the misunderstanding. However, on a MATH forum, one would expect symbols to take their math definitions and not code form. Next time, please clarify by using mathematics in math forums, even if only in brackets. In case you were unaware, it is unicode 2260. Sorry for the confusion.[/QUOTE]

it's not just a math forum though it's got a programming section in the main forums. Anyways I caught the difference just took me a while. The main problem i had was as described x"! =" is likely factorial, x!= you can't tell without doing math, and x" !=", is x is not equal to.

[QUOTE=science_man_88;258335]it's not just a math forum though it's got a programming section in the main forums. Anyways I caught the difference just took me a while. The main problem i had was as described x"! =" is likely factorial, x!= you can't tell without doing math, and x" !=", is x is not equal to.[/QUOTE]

of course x! = and [TEX]\ne[/TEX] solves such arguments.