Question

3 integers are in an arithmetic progression. their product is prime. what is the sum of their squares?

Answer 1

The sum of their squares is (x)^2 + (x-d)^2 + (x-2d)^2.

According to the statement

we have given that the 3 integers are in an arithmetic progression and their product is prime.

And we have to find the sum of their squares.

So, let the three integers which is A, B, C.

And according to the arithmetic progression the

A = x and B = x-d, C = x-2d. here d is the common difference and x is a 1st term.

So, there product is prime

then

(x) (x-d) (x-2d) = c -(1)

And

their sum of square is

(x)^2 + (x-d)^2 + (x-2d)^2 - (2)

So, The sum of their squares is (x)^2 + (x-d)^2 + (x-2d)^2.

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