Today is 8 day of the year.
8 = 2^3.
8 = 3 + 5, i.e. the sum of two consecutive primes.
8 is a member of just one primitive, Pthagorean triplet, namely (8, 15, 17).
8 is the third cake number.
8 is the sixth Fibonacci number.
It is conjectured that the difference between the squares of any two odd primes is a multiple of 8.
This is no longer a conjecture since it is easy to prove, as follows:
Consider an odd number, x = 2k + 1 for some k >= 0
Then x^2 = (2k+1)^2 = (2k + 1)(2k + 1) = 4k^2 + 4k + 1 = 4(k^2 +k) + 1
Consider k^2 + k
If k is odd then k^2 is odd and, therefore, k^2 + k is even.
If k is even then k^2 is even and, therefore k^2 + k is even.
Thus k^2 + k = 2l for some l >= 0.
Which means x^2 = 4.2l + 1 = 8l + 1.
Assume that we have another odd number, y,then we can safely assume that y^2 = 2m + 1 for some m >= 0.
This the difference between the two odd numbers squared, assuming x > y,
x^2 - y^2 = (8l + 1) - (8m + 1) = 8l + 1 - 8m - 1 = 8l - 8m = 8(l - m).
Thus we have shown that the difference between two odd numbers squared is a multiple of 8. Since the odd prime numbers are a subset of the odd numbers then this result holds for them as well.
For a conjecture that uses this result see Pe's conjecture.