Nice.
And just across the river is HoBroken.
Sounds like baby mama didn't give a shit her hubby just died.
Z and his cabinet are literally TV actors, so…
If we were really screwed we'd all be dead already.
Now that you know who's screwing you, how much longer are you going to just sit on your ass and cry about it?
I don't swing your way, homo.
Only holding down the shovels so they don't float away.
LAGRANGE’S FOUR SQUARE THEOREM
Euler’s four squares identity. For any numbers a, b, c, d, w, x, y, z
(a2 + b2 + c2 + d2)(w2 + x2 + y2 + z2) = (aw − bx − cy − dz)2+
(ax + bw + cz − dy)2 + (ay + cw + dx − bz)2 + (az + dw + by − cx)2.
Lagrange’s Theorem. Every natural number is the sum of four squares.
Proof. In view of Euler’s identity and 12 + 12 = 2, it suffices to prove that every odd
prime is such a sum.
Lemma 1. If n is even and is a sum of four squares, then so is n
2 .
Proof of Lemma 1. When n = a2 + b2 + c2 + d2 is even, an even number of the squares
will be odd. and so the a, b, c, d can be rearranged so that a, b have the same parity and
so do c, d. Thus n
2 = ( a+b
2
)2 + ( a−b
2
)2 + ( c+d
2
)2 + ( c−d
2
)2.
Lemma 2. If p is an odd prime, then there are integers a, b, c, d and an m so that
0 < a2 + b2 + c2 + d2 = mp < p2
2 .
Proof of Lemma 2. The p+1
2 numbers 02, 12 . . . , ( p−1
2
)2 are pairwise incongruent modulo
p. Hence, by a box argument there are u, v such that u2 ≡ −v2 − 1 (mod p) and 0 <
u2 + v2 + 1 ≤ p2−2p+3
2 .
By Lemma 2 there is an integer m with 0 < m < p so that for some a, b, c, d we have
a2 + b2 + c2 + d2 = mp
and we may suppose that m is chosen minimally. Moreover, by Lemma 1 we may suppose
that m is odd. If m = 1, then we are done. Suppose m 1. If m were to divide
each of a, b, c, d, then we would have m|p contradicting m < p. Choose w, x, y, z so
that w ≡ a (mod m), |w| ≤ m−1
2 , x ≡ −b (mod m), |x| ≤ m−1
2 , y ≡ −c (mod m),
|y| ≤ m−1
2 , z ≡ −d (mod m), |z| ≤ m−1
2 , and then not all of w, x, y, z can be 0. Moreover
w2 +x2 +y2 +z2 ≡ 0 (mod m) and so 0 < w2 +x2 +y2 +z2 = mn ≤ 4 ( m−1
2
)2 = (m−1)2.
Thus 0 < n < m. Now aw−bx−cy−dz ≡ a2+b2+c2+d2 ≡ 0 (mod m), ax+bw+cz−dy ≡
−ab + ab − cd + dc ≡ 0 (mod m), ay + cw + dx − bz ≡ −ac + ac − db + db ≡ 0 (mod m),
az + dw + by − cx ≡ −ad + ad − bc + bc ≡ 0 (mod m). By Euler’s identity m2np is the
sum of four squares and each of the squares is divisible by m2. Hence np is the sum of
four squares. But n < m contradicting the minimality of m.
Mornin', Pig.