Question #b6464

1 Answer
Jan 28, 2017

Answer:

See below.

Explanation:

if #p# is an odd integer, #p ge 3# then

#(p,(p^2-1)/2,(p^2+1)/2)# is a primitive Pythagorean triple.

You can verify easily that #p^2 + ((p^2 - 1)/2)^2=((p^2+1)/2)^2#

then making #a=(p^2-1)/2# and #b=(p^2+1)/2# it follows

#2(p+a+1)=2(p+(p^2-1)/2+1)=2p+p^2-1+2=(p+1)^2#

Now being a Pythagorean triple, exactly one of #p,a# is divisible by #3# but #p# is prime #> 3# then #a# is divisible by #3#. Also #(p^2-1)# is divisible by #4#. The multiplicity by #12# attached to #a# is left as an exercise.