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.
BurningFrog|1 year ago
odd + odd is even, as is even + even.
Spivak|1 year ago
All the symbol soup proofs aren't wrong but I don't think they satisfyingly explain the why.
crabbone|1 year ago
MichaelRo|1 year ago
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.
dleeftink|1 year ago
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?
cherryteastain|1 year ago
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.
svat|1 year ago
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).
abnry|1 year ago
deruta|1 year ago
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))
dh2022|1 year ago