| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1719.1 | | SSAG::LARY | Laughter & hope & a sock in the eye | Sat Feb 27 1993 03:00 | 20 |
| With a little algebra, the answer appears to be equivalent to finding a,b such
that
7 does not divide ab(a+b)
7^3 divides a^2 + ab + b^2
and this sets some constraints on a and b, namely that they must be congruent
to one of the following pairs of numbers mod 7:
a b
--- ---
1 2
1 4
2 4
3 5
3 6
5 6
But I'm going skiing in the morning and I can't go on like this...
|
| 1719.2 | As I was dropping off... | SSAG::LARY | Laughter & hope & a sock in the eye | Sat Feb 27 1993 03:16 | 2 |
| a = 1 and b = 343k + 324 is one solution...
|
| 1719.4 | an answer | HERON::BUCHANAN | The was not found. | Wed Mar 03 1993 04:07 | 12 |
| It's simpler than I'd thought.
(a+b)^7 - a^7 - b^7 = 7ab(a�+ab+b�)�
Therefore:
7^7 | (a+b)^7 - a^7 - b^7
=> 7^3 | a� + ab + b�
=> 7^3 | a� - b�
=> 7^3 | (a/b)� - 1
=> a/b == 18 or (18� =) -19 mod 343
and that's it.
|
| 1719.5 | | STAR::ABBASI | i think iam psychic | Wed Mar 03 1993 08:07 | 17 |
| ref .4
> (a+b)^7 - a^7 - b^7 = 7ab(a�+ab+b�)�
humm? that can't be, 'cause the right hand side will generate a term
with at most a^5 and b^5 in them.
while the left hand side will have an a^6 and b^6 terms in them!
did i , did i, finnd a mistake by Andrew!
if so, i want to save this note and decorate too.
\bye
\nasser
ps. in case iam wrong, please note i have not had my morning coffe yet.
|
| 1719.6 | fooey | HERON::BUCHANAN | The was not found. | Wed Mar 03 1993 08:43 | 18 |
| >> (a+b)^7 - a^7 - b^7 = 7ab(a�+ab+b�)�
> humm? that can't be
Oops.
maple
factor ((a+b)^7 - a^7 - b^7 );
2 2 2
7 a b (a + b) (a + a b + b )
Drat, I forgot to remember one factor. It doesn't affect the proof, though.
> did i , did i, finnd a mistake by Andrew!
Thanks for the flattery, but my errors are all too common :-)
Andrew.
|
| 1719.7 | | AUSSIE::GARSON | | Wed Mar 03 1993 20:53 | 9 |
| re .4
> => 7^3 | (a/b)� - 1
>
> => a/b == 18 or (18� =) -19 mod 343
Is there any easy way to get from 7� | x� - 1 to x = 18 as one solution?
Also, you don't mention why you eliminate the case a/b == 1 (mod 343).
|
| 1719.8 | | SSAG::LARY | Laughter & hope & a sock in the eye | Thu Mar 04 1993 14:09 | 19 |
| Re .-1: In the derivation from .4,
...
=> 7^3 | a� + ab + b�
=> 7^3 | a� - b�
=> 7^3 | (a/b)� - 1
That second step isn't reversible if (a-b) is divisible by 7. That's why
a = b (mod 343) isn't a solution, and neither are some of the cases where
a = 18b (mod 343) and a = -19b (mod 343). For example, (a,b) = (7,126) is
obviously not a solution since 7 | ab(a+b). Its simple to show that if
a = b (mod 7) then, given the other constraints, a = 0 (mod 7) and its not a
solution.
So, the complete set of solutions therefore is given by:
a = 18b (mod 343) or a = -19b (mod 343), AND a <> b (mod 7)
(As for extracting the cube roots of 1 (mod 343), I dunno, but an exhaustive
search takes minutes to code and milliseconds to run...)
|
| 1719.9 | | HERON::BUCHANAN | The was not found. | Fri Mar 05 1993 03:48 | 6 |
| The point of bringing in the (a/b)�-1 business is to show why there
are two solutions and each is the square of the other, mod 343. When it comes
to *finding* the solutions, I spose you can take a�+ab+b� and complete the
square mod 343. This just leaves a square root mod 343 to find. By machine
I'd numbercrunch, by hand I'd solve mod 7, then mod 49, finally mod 343.
Andrew.
|