Where do those undergraduate divisibility problems come from?
grossack.site
grossack.site
The OP blog post here actually links to / points out there's an Wikipedia article on integer-valued polynomials: https://en.wikipedia.org/wiki/Integer-valued_polynomial
Among other things, as mentioned, “every integer-valued polynomial can be written as an integer linear combination of binomial coefficients” [in exactly one way].
The conjecture in the OP post, that every polynomial everywhere divisible by k counts {something}, is intriguing, and I wouldn't be surprised if it were true.
There is a genre of undergraduate polynomial divisibility problems which look like this: Show that f(n) is divisible by some integer k.
These problems often appear to be (elementary) number theory problems. However, often there is a rather elegant proof associated with them which is based on combinatorics.
The crux of this proof is that the polynomial counts the number of equivalence classes of a certain kind.
This is closely related to https://en.wikipedia.org/wiki/Burnside%27s_lemma
The question at the end of the post is whether _all_ such problems must come this way
Exercise left to the reader:
Prove 7*n^3 + n is divisible by 2
odd + odd is even, as is even + even.
All the symbol soup proofs aren't wrong but I don't think they satisfyingly explain the why.
For divisibility by two, there are only two cases. So if n is 1, then n³ is 1, and if n is 0, n³ is 0. 0+0 = 0; 1+1 = 0; and this completes the proof.
I am not actually sure that doing a prime factorization on 7n³ for unknown n is easier than knowing that 1³ = 1.
If n is even, we can choose some m such that n = 2m, and p(n) = p(2m) = 7 * 8m^3 + 2m = 2 * (7 * 4m^3 + m), which is divisible by 2 since we could factor out the 2 at the start.
If n is odd, similarly we can say n = 2m + 1. p(2m) = 7 * (2m + 1)^3 + (2m + 1) = 56m^3 + 84m^2 + 44m + 8 = 2 * (28m^3 + 42m^2 + 22m + 4), which is also divisible by 2 per the 2 at the start.
Took a look on it, seems like a highly particular / specialized area of mathematics. It's like computer science, can't know them all. If you work all day with some area, say compilers or databases or financial software or what else, you'd be a whizz at it while it's unreasonable to expect someone from a different domain be able of more than a superficial understanding of what you write.
I'm pretty good at math but like with computers, I don't have the compulsion to dive deep into an unfamiliar domain just for the sake of it. So commenting on the article: cool, now I know how these problems are formed and in the very unlikely domain I'll need to produce one, I know where to look. Likely this will never happen, though.
As a non-mathy, I'm interested in whether the idea that being good/able to provide proofs in one area, automagically makes one proficient in another is customary in the field or rejected quite early on when choosing a math specialisation?
- There are skills that carry over; these are usually known by the name "mathematical maturity" https://en.wikipedia.org/w/index.php?title=Mathematical_matu...
- There is a story/legend told about Erdős, where he was so good at problem-solving/proofs that he once solved a problem in another area after asking for the definitions of the terms in the problem. (The fact that this story is told illustrates that it is not commonplace.)
Like take for instance financial mathematics where I had some special interest, it's totally oblivious to areas such as geometry or number theory. I never had to figure out if a polynomial is divisible by 6 for instance :)
Like computer science, there's the common algorithms stuff but being an expert in web development doesn't help you much in writing high frequency trading server code, and the other way around.
1 + 2 + ... + n = n(n+1)/2
2 divides n(n+1), n(n+1) = 2m
7 * n^3 + n =
2*(3*n^3 + n) + n^3 - n =
2*(3*n^3 + n) + n(n+1)(n-1) =
2*(3*n^3 + n + m(n-1))
As the post mentions in passing, the integer-valued polynomials are completely characterized by the property that when written as a sum of {c_i (x choose i)}, all the coefficients c_i are integers. I imagine this is where most of the exercises actually come from. For example, using [3 1 4 1 5 9], the polynomial {3 + 1·x + 4·x(x-1)/2 + 1·x(x-1)(x-2)/6 + 5·x(x-1)(x-2)(x-3)/24 + 9·x(x-1)(x-2)(x-3)(x-4)/120} simplifies to 1/120 (9x^5 - 65x^4 + 185x^3 + 5x^2 - 14x + 360), so you could use it to generate exercises like:
- Prove that 9x^5 - 65x^4 + 185x^3 + 5x^2 - 14x + 360 is always a multiple of 120
(or 5, or any divisor of 120).
because it's not true (simply insert 1, 2, 4 or 5)
the polynomial is =0 mod2 and =0 mod3 so its =0 mod6
n^6 + n^3 + 2n^2 + 2n (mod 2) = n^6 + n^3 + 0 + 0 = n^3(n^3+1) = 0*1 or 1*0 = 0
because consecutive numbers are even then odd then even ....
for mod3 you can make a table
you could also factor the polynomial and see the solution easily
n(n+1)(n^2-2n+2)(n^2+n+1)
Thinking about it I guess MSN messenger and My Space also allowed/encouraged font shenanigans? My memory falters