Conference rusure::math

1861.0. "why does 10001 = 73 * 137 ?" by HANNAH::OSMAN (see HANNAH::IGLOO$:[OSMAN]ERIC.VT240) Mon Mar 21 1994 18:08

100001 is divisible by 11.  One way of saying "why" this is so is that
100001 is

	x^5 + 1   (x = 10)

and by polynomial division, we see that

	x^5 + 1 = ( x + 1 ) ( x^4 - x^3 + x^2 - x + 1)

In our case x + 1 = 11, so that's "why" 100001 is divisible by 11.

1001 is divisible by 11 because 1001 is of the form x^3+1 for x = 10.

Similarly, 1000001, being x^3+1 is divisible by 101 "because" polynomial
division shows that x^3 + 1 is divisible by x+1, and x is 100 in our case.


Is there always such a "why" for any factorization ?  Can some similar
explanation be found for "why" 10001 = 73 * 137.



We could do one backwards.  Out of a hat, I can pick

	(x+1)(x^2-3x+3)

Now I can just plug in x=10.

	11 * 100-30+3 = 11 * 73 = 803

So, armed with this knowledge, if someone happened to ask me:

	Why is 803 divisible by 11 ?

I could say

	Oh it's obvious because 803 is of the form (x+1)(x^2-3x-3) for x=10 !!


But how does one work from the factorization back to the polynomials ?

We certainly don't have to stick to the number 10.  Maybe that's the answer.
Maybe 10001 is of some simple polynomial form for some number other than 10 ?

But what's the "best" number to pick to illustrate the 73 * 137 ?

/Eric

    Well, yeah, a dumb solution....
    For all odd products we have a*b=(a+b)/2 * (a-b)/2, or offhand we have
    10001=(x+32)(x-32) for x=105. Simplify to 10001=(x-y**5)(x+y**5) for
    x=105,y=2 and your public is really stunned. I hope this helps ( Though
    fear it doesn't. ( How about (5x-y**5)(5x+y**5) =;-)
    
    One time Math Olympiad winner Michel

    Re .1 I mean a*b=((a+b)/2)**2-((a-b)/2)**2 , where both (a+b)/2 and (a-b)/2
    are integer if a and b are both odd or both even. Let (a+b)/2=x and
    (a-b)/2=y, this reads as x**2-y**2, or (x+y)(x-y) as a polynomial.

    It is a bit like 'Anything goes what you can get away with'. For
    example take 2.2.3.3.7.13 as an example. Choose to factor in 
    2.13 and 2.3.3.7, gives 26.126, or (5^2+1)(5^3+1), or (x^2+1)(x^3+1)
    for x=5.
    using 2.3.3, 2.7,13 gives 18.14.13, or x(x+1)(x+5) for x=13
    What is still simple in your eyes? x(x+2)(x^2+3) for x=7
    
    In general, given a number a1.a2...an, partition in a selected number
    of factors and choose a number x such that
    x is close to a1, x^2 close to a2 etc.   That will give your polynomial
    in a simple form. Most impressive with primes of course.
    7*37=(x+1)(x^2+1)
    
    Hope this gives a full overview, and answers your question.
    Michel