Verified answer

Theorem: the number of edges in a graph is the sum of the degrees divided by 2.

Proof: Every edge hits two vertices, so by adding up the degrees we count every edge twice.

Now clearly the number of edges in a graph has to be an integer. There is no such thing as a graph with 5.79 edges. So (1+2+..+n)/2 should better be an integer. But 1+2+..+n = n(n+1)/2. So we need

n(n+1)/4 to be an integer. If n is odd, that means n+1 is divisible by 4, ie n ≡ 3 (mod 4). If n is even then n+1 is odd, so 4 divides n, ie n ≡ 0 (mod 4). So we have either n ≡ 0 (mod 4) or n ≡ 3 (mod 4).

Hope this helps:)

i imagine induction is the right thanks to bypass. Step a million: Set n=0. Plug in to get f(0) = 0+0+0-0 = 0, an integer. Step 2: assume f(x) is an integer for all x = 0, a million, ..., n. Now try x = n+a million. f(n+a million) = (n+a million)^5/5 + (n+a million)^4/2 + (n+a million)^3/3 - (n+a million)/30. i visit set up the words by technique of power of n at the same time as doing the binomial theorem in my head: f(n+a million) = n^5 * (a million/5) + n^4 * (5/5 + a million/2) + n^3 * (10/5 + 4/2 + a million/3) + n^2 * (10/5 + 6/2 + 3/3) + n * (5/5 + 4/2 + 3/3 - a million/30) + (a million/5 + a million/2 + a million/3 - a million/30). Recombining words in a fashion to resemble f(n): f(n+a million) = (n^5/5 + n^4/2 + n^3/3 - n/30) + n^4 + 4*n^3 + 6n^2 + 4n + a million = f(n) + n^4 + 4*n^3 + 6n^2 + 4n + a million, that is an integer.