This one regarding prime numbers is certainly not as tough as it may look ?

Question

Prove that where
a^2 +b^2 +c^2 ........ = z^2 ;
a, b, c........and z, all of them can not be prime numbers > 3, unless at least all alphabet letters are used up for the equation - in other words, LHS ought to have at least 25 appropriate prime numbers

Answer 1

Fun problem :) It is not true, of course, because we can solve a^2 = z^2, assuming that two equal prime numbers are allowed -- but I am sure you want to rule that out.

The prime numbers are odd, and therefore can be written as x = 2y + 1. Therefore, their squares are of the form
x^2 = 4(y^2 + y) + 1
which is a multiple of 8 plus 1. It follows that the number of terms at the LHS is a multiple of 8 plus 1.

The prime numbers are greater then 3, and therefore no multiples of three: x = (3y +- 1), and therefore their squares
x^2 = 3(3y^2 +- 2y) + 1 are multiples of 3 plus 1. It follows that the number of terms at the LHS is a multiple of 3 plus 1.

Consequently, the number of terms at the left hand side must be a multiple of 24 plus 1 -- at least 25, if we rule out a single term.

Answer 2

well the answer is correct if i was your teacher id give you a A+ good job