>>20247278Pic was one
The other is
For a given integer k, the triplets:
(a, b, k), (a, k, b), (k, a, b) will remove 2 numbers
(a, k, k), (k, a, k), (k, k, a) will remove 1 number
(k, k, k) will remove 0 numbers
(where a, b < k)
Amount of numbers removed is
= 2*(remove 2 nums) + 1*(remove 1 num)
= 2*3(k-1)^2 + 1*3(k-1)
= 6k^2-9k+3
The number remaining is
(num numbers) - sum_(k=1 to n)
= 3n^3 - sum(k=1 to n)(6k^2-9k+3)
= (algebra im too lazy to type)
= n(n+1)(2n+1)/2
For the expression to be a square, two must be perfect squares and one must be twice a perf square
or, their products must be twice a square, both cases is represented like this
n = r^2*k1
2n+1 = s^2*k2
n+1 = t^2*k3
where k1,k2,k3 are squarefree and k1*k2*k3=2x^2
two squarefree numers' products are a square iff they are equal. since k1 != k3 because n and n+1 have no
common facors. however, neither can k1 = k2, because 2n + 1 can't have a common factor to 2n, which has n as a factor. neither can k3 = k2, since 2n + 1 and 2n+2 have no common factors, and 2n + 2 = 2(n+1) has n + 1 has a factor. this leaves only the case that k1 = k2 = k3
k1 = k2 = k3 = 1 (oops typo in last tweet), but that means that
n(n+1)(2n+1)/2 = (rst)^2/2
if rst is even, then (rst)^2/2 = 4k^2/2 = 2k^2, which is not a perfect square, if rst is odd, then (rst)^2/2 is not an integer. no values exist, therefore it is never a perf square. qed
