How many Pythagorean triples, primitive or not, contain a 28?

x^2 + 28^2 = z^2

784 = z^2 - x^2 = (z - x)(z + x)

(z-x) ··· (z+x) ··· z ··· x

····1 ······ 784 ···

····2 ······ 392 195 197

····4 ······ 196 ··96 100

····7 ······ 112···

····8 ········ 98 ···45 53

···14 ········ 56·· 21 35

···16 ········ 49

···28 ········ 28

195² + 28² = 197²

96² + 28² = 100²

45² + 28² = 53²

21² + 28² = 35²

I did this by looking at cases in Excel.

There are no combinations of integers where a^2 + b^2 = 28^2, so 28 can't be the hypotenuse.

Therefore 28 has to be one of the legs. If the other leg is n, then it must be less than 391, because if it's 391 the hypotenuse is 392.0012755, but if it's 392 the hypotenuse is only 392.9987277, which is less than 1 greater than the other leg. The difference just gets smaller from there, so any integer leg must be < 391. Checking them all, we find the following combinations yield integers for the two sides and the hypotenuse

28 21 35

28 45 53

28 96 100

28 195 197

x = 2mn

y = m^2 - n^2

z = m^2 + n^2

m > 0

n > 0

m > n

"m or n" must be even

m = 7 and n = 2; 28, 43, and 51 ( primitive)

m = 2 and n = 1; 4, 3, and 5: this turns into 28, 21, and 35