7*n^3 is even when n is even and odd otherwise.

odd + odd is even, as is even + even.

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.

All of these "always divisible by n" proofs are asking you to solve them case by case in modular arithmetic.

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.

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?

It is not typical that being good in one area of mathematics makes one proficient in another — mathematics has a lot of depth, and to reach the frontier in any specialization requires years of study. However:

- 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.)

Well there is common stuff that one must be familiar with to be proficient and marginal stuff that you can get by without knowing because chances is you're not going to need it.

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.

Given p(n) = 7n^3 + n:

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.

> but not to write the exercises

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).

7n^3 +n (mod 2) = 1 n^3 + n = n + n = 2*n = 0*n = 0

And a proof by "counting something":

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))

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.