That blog post is specifically about the question about whether a polynomial taking integer values on positive integers (or any sufficiently large subset thereof) necessarily is an integer-valued polynomial (the answer is yes).
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.
The easy way of seeing the first part is to do the prime factorization. The 7 doesn't matter since it's prime. If n has a 2 in its factorization it now has 2^3. But if it doesn't have a 2 it won't suddenly acquire one.
All the symbol soup proofs aren't wrong but I don't think they satisfyingly explain the why.
I vote this the best proof. All you need to know to understand it is to know how multiplication, addition and exponentiation works. You could probably show this to a child in a sixth grade or so, and have them understand it. This is really good!
>> Apparently I'm good enough at math to do the proofs, but not to write the exercises.
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.
> seems like a highly particular / specialized area of mathematics. It's like computer science, can't know them all
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?
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.
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
n = 0 or 1 modulo 2. So we have to check out only two cases module 2, and these cases trivial. To prove the problem note that 6 = 2 * 3 and then is trivial to see that the polynomial is=0 modulo 2 if n=0,1 modulo 6and then check it is =0 modulo 3 for n=0,1,2 modulo 3 and you are done.
You're the second commenter, so far, to mention exclamation marks. What do they mean to you that would bother you so much to point it out, or anyone for that matter? I haven't even noticed them until I read the comments here on hn.
Oh this is nothing. One of my colleagues does that and adds random colour changes, underlines and font face changes. It's like working with a serial killer.
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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.
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?
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.
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).
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))
because it's not true (simply insert 1, 2, 4 or 5)
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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)
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