OFFSET
11,1
COMMENTS
A combinatorial meaning is that if there are x red socks and y blue socks, then there are 2(mn)^2 more ways to choose a two-colored pair than a single-colored pair, where m and n are the unique relatively prime integers such that a = m^2 - n^2, b = 2mn, c = m^2 + n^2 and m != n (mod 2). The proof is just that xy - binomial(x,2) - binomial(y,2) = (c^2 - a^2)/2 = (b^2)/2, when xy = (c^2 - a)/2 (c^2 + a)/2, and (a,b,c) is a PPT.
Since there are c^2 socks in total, it is also possible to partition them into binomial(c+1,2) red socks and binomial(c,2) blue socks, whence there are as many ways of choosing a single-colored pair or a two-colored pair, because the equation binomial(c+1,2) binomial(c,2) = binomial(binomial(c+1,2),2) + binomial(binomial(c,2),2) is equivalent with the true equation (binomial(c+1,2) - (c,2))^2 = binomial(c+1,2) + binomial(c,2) = c^2.
There is a special relationship between xy and binomial(c+1,2) binomial(c,2), when (a,b,c) is a PPT, namely that xy - binomial(c+1,2) binomial(c,2) = (mn)^2. This is somewhat easy to deduce, because xy - binomial(binomial(c+1,2),2) - binomial(binomial(c,2),2) = (c^2 - a^2)/4 = (b^2)/4, when xy = (c^2 - a)/2 (c^2 + a)/2. Thus, in the partition to x and y, the possible ways of choosing a two-colored pair exceed binomial(c+1,2) binomial(c,2) by (mn)^2, and the possible ways of choosing a single-colored pair undercut it by (mn)^2.
A positive integer can be partitioned into red and blue members with as many ways of choosing a single- or a two-colored pair if and only if it is a perfect square. The 'if' part was already shown with the socks above. The 'only if' part of the proof is as follows: there are A + B socks, of which A are red and B are blue and B > A, then B = A + K for some positive integer K. Suppose that (A-B)^2 = A + B, that is, K^2 = 2A + K, equivalently K^2 - K = 2A, equivalently A = binomial(K,2). Thus B = A + K = (K^2 - K)/2 + K = binomial(K+1,2). Therefore, A + B = binomial(K+1,2) + binomial(K,2) = K^2.
Furthermore, if (a,b,c) is a PPT, and the square c^2 is partitioned into the squares a^2 and b^2, then there are more ways to choose a single-colored pair than a two-colored pair, because (a^2 - b^2)^2 = (a-b)^2*(a+b)^2 >= (a+b)^2 > c^2 = a^2 + b^2.
Also, if there are equal numbers of both colors, then there are always fewer ways to choose a single-color pair, no matter if the total number of socks is a square or not. The proof is that A^2 > 2 binomial(A,2) for any positive integer A.
The formula for PPT using integers m and n is a well-known result. It can be proved using elementary number theory. The proof is omitted here.
LINKS
Wikipedia, Pythagorean triple.
FORMULA
x = (c^2 +- a)/2, pick the odd number, one is guaranteed to be odd and the other one is even.
Proof. Since c^2 and a are both odd numbers, and c^2 > a, it follows that c^2 +- a = 2x and c^2 -+ a = 2y, for some positive integers x and y. Thus 2c^2 = 2(x+y), whence c^2 = x + y, making x and y of opposite parity, and 2a = +-2x -+ 2y. Thus +-x = a +- y.
CROSSREFS
KEYWORD
nonn
AUTHOR
Laura Jokinen, Oct 02 2022
STATUS
approved