Sets of Numbers!
Let be the set of all numbers of the form , where is a natural number. Find the sum of all positive integral values of such that the product also belongs to for all natural numbers .
The answer is 1.
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1 solution
[a(n)][a(n+k)] = [n^2 + n + 1][(n+k)^2 + (n+k) +1] = n^4 + 2(k+1)(n^3) + (k^2 + 3k +3)(n^2) + (k^2 +3k + 2)n + k(k+1) + 1
This product is also a member of set S, so it equals a(m) for some positive integer m.
Thus, a(m) = m^2 + m + 1 = n^4 + 2(k+1)(n^3) + (k^2 + 3k +3)(n^2) + (k^2 +3k + 2)n + k(k+1) + 1.
m^2 + m = n^4 + 2(k+1)(n^3) + (k^2 + 3k +3)(n^2) + (k^2 +3k + 2)n + k(k+1)
m(m+1) = n^4 + 2(k+1)(n^3) + (k^2 + 3k +3)(n^2) + (k^2 +3k + 2)n + k(k+1)
So if we were to factorize this polynomial as factors m and m+1, the degree of m is 2 as the degree of the polynomial is 4. For the n-term in m, the coefficient is (k+1) as the coefficient of n^3 in the polynomial is 2(k+1), and the final term in m is k (which will be multiplied to (k+1) in m+1), for the last term in the polynomial is k(k+1):
m = n^2 + (k+1)n + k; m+1 = n^2 + (k+1)n + (k+1)
m(m+1) = (n^2 + (k+1)n + k)(n^2 + (k+1)n + (k+1)) = n^4 + 2(k+1)(n^3) + (k^2 + 3k +3)(n^2) + (k^2 +3k + 2)n + k(k+1)
All the terms are cancelled out except for the n-terms, where we are left with:
[(k+1)^2 + k^2 + k]n = (k^2 +3k + 2)n
Solving for possible k, 2k^2 + 3k + 1 = k^2 +3k + 2; k^2 = 1.
Thus, k = 1 only. The sum of such k is, therefore, 1.