More on integer valued polynomials: <a href="https://cameroncounts.wordpress.com/2017/01/31/polynomials-taking-integer-values/" rel="nofollow">https://cameroncounts.wordpress.com/2017/01/31/polynomials-t...</a>
TLDR Summary:<p>There is a genre of undergraduate polynomial divisibility problems which look like this: Show that f(n) is divisible by some integer k.<p>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.<p>The crux of this proof is that the polynomial counts the number of equivalence classes of a certain kind.<p>This is closely related to <a href="https://en.wikipedia.org/wiki/Burnside%27s_lemma" rel="nofollow">https://en.wikipedia.org/wiki/Burnside%27s_lemma</a><p>The question at the end of the post is whether _all_ such problems must come this way
Well I was curious, but there's a lot there I didn't understand. Apparently I'm good enough at math to do the proofs, but not to write the exercises.<p>Exercise left to the reader:<p>Prove 7*n^3 + n is divisible by 2
i couldnt come up with a proof for the initial problem (n^6+n^3+2n^2 is a multiple of 6 for every n)<p>because it's not true (simply insert 1, 2, 4 or 5)
i just computed the solution mod 2 and mod 3 a la chinese remainder theorem<p>the polynomial is =0 mod2 and =0 mod3 so its =0 mod6<p>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<p>because consecutive numbers are even then odd then even ....<p>for mod3 you can make a table<p>you could also factor the polynomial and see the solution easily<p>n(n+1)(n^2-2n+2)(n^2+n+1)