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.

Update:

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.

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