### Monday, September 27, 2010

## JSH: Attacking the puzzle

Rather than do yet ANOTHER explanation of an issue in number theory, I presented the issue as a puzzle. Some posters have replied in that thread, and those who tried to solve the puzzle failed massively, but protested when I noted that they'd failed.

So here is an attack on the puzzle using the knee-jerk responses that are WRONG. So it's easily to massively fail against this thing.

Here again is the puzzle, and I'll then attack it just a bit.

Try in the ring of algebraic integers to use:

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

where g_1(0) = g_2(0) = 0, and one of the f's equals 0 at x=0, as well, while P(x) is a quadratic with integer coefficients.

You can find cases in the ring of algebraic integers where NEITHER of the f's can have 9 as a factor, for certain values of x, so you wander off the to the complex plane—because you've a VERY SMART math student—so you can see exactly what is happening.

There you note that:

(9g_1(x) + 9)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

follows trivially by the distributive property so FOR THAT CASE the ring of algebraic integers presumably would have 9 as a factor for all x, but can be shown to NOT have 9 as a factor as noted above.

Being an EXTREMELY INTELLIGENT math student you find this puzzling, so you consider it until it makes sense.

First off, let's attack the idea of going to the complex plane as a divisibility issue as it DOES NOT MATTER on the complex plane.

Well, note that the result HOLDS for polynomial functions for the f's and g's. So with all integers—valid on the complex plane—you do indeed find that one of the f's has 9 as a factor for all x.

Notice that following that on the complex plane does not remove the result.

The complex plane INCLUDES the ring of integers, so it can look at something happening in that ring easily.

If divisibility were the issue, it would emerge with integers as well, and everything would be divisible by 9. But one of the f's is not in general divisible by 9, so no, divisibility as the issue is a massive fail for this puzzle.

One thing that I think is fascinating as a bizarre issue is that some posters seem lost on the concept that you can use the complex field to simply look at something happening in a ring, like the ring of integers. May seem like overkill, but you can do it!

So making that mistake actually also says something about your sophistication as a math student, as presumably a fairly astute math student would realize that trivially.

Another poster actually said, there was no puzzle!!! Clearly that one is a massive fail as my easy counter is to explain it then. But his answer was to try and claim divisibility, so another massive fail. Claiming there is no puzzle but not being able to answer correctly is not an answer.

Can you now figure out the puzzle?

What others issue need I address? This thread can be for attacking the puzzle, the validity of it as a puzzle, or if you want, try to give an answer. I assure you it's not a trivial task to answer it correctly.

So here is an attack on the puzzle using the knee-jerk responses that are WRONG. So it's easily to massively fail against this thing.

Here again is the puzzle, and I'll then attack it just a bit.

Try in the ring of algebraic integers to use:

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

where g_1(0) = g_2(0) = 0, and one of the f's equals 0 at x=0, as well, while P(x) is a quadratic with integer coefficients.

You can find cases in the ring of algebraic integers where NEITHER of the f's can have 9 as a factor, for certain values of x, so you wander off the to the complex plane—because you've a VERY SMART math student—so you can see exactly what is happening.

There you note that:

(9g_1(x) + 9)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

follows trivially by the distributive property so FOR THAT CASE the ring of algebraic integers presumably would have 9 as a factor for all x, but can be shown to NOT have 9 as a factor as noted above.

Being an EXTREMELY INTELLIGENT math student you find this puzzling, so you consider it until it makes sense.

First off, let's attack the idea of going to the complex plane as a divisibility issue as it DOES NOT MATTER on the complex plane.

Well, note that the result HOLDS for polynomial functions for the f's and g's. So with all integers—valid on the complex plane—you do indeed find that one of the f's has 9 as a factor for all x.

Notice that following that on the complex plane does not remove the result.

The complex plane INCLUDES the ring of integers, so it can look at something happening in that ring easily.

If divisibility were the issue, it would emerge with integers as well, and everything would be divisible by 9. But one of the f's is not in general divisible by 9, so no, divisibility as the issue is a massive fail for this puzzle.

One thing that I think is fascinating as a bizarre issue is that some posters seem lost on the concept that you can use the complex field to simply look at something happening in a ring, like the ring of integers. May seem like overkill, but you can do it!

So making that mistake actually also says something about your sophistication as a math student, as presumably a fairly astute math student would realize that trivially.

Another poster actually said, there was no puzzle!!! Clearly that one is a massive fail as my easy counter is to explain it then. But his answer was to try and claim divisibility, so another massive fail. Claiming there is no puzzle but not being able to answer correctly is not an answer.

Can you now figure out the puzzle?

What others issue need I address? This thread can be for attacking the puzzle, the validity of it as a puzzle, or if you want, try to give an answer. I assure you it's not a trivial task to answer it correctly.