Question

Prove that if an integer n is the sum of two squares (n = a2+ b2 for a, b Z) then n = 4q or n= 4q+1 or n= 4q + 2 for some qЄ Z. Deduce that 1234567 cannot be written as the sum of two squares.

Answer 1

The number 1234567 cannot be written as a sum of two squares.

Let q = x^2+y^2,

Let a and b be two even numbers, such that

a= 2x and b = 2y

then

n = (2x)^2 + (2y)^2 = 4 (x^2+y^2)

implies n = 4q

Let a and b be two odd numbers, such that

a = 2x+1 and b = 2y+1

then

n = (2x+1)^2 + (2y+1)^2 = 4x^2+4y^2+1 = 4q+2

Let a be an even number and b be an odd number, such that

a= 2x and b = 2y+1

then

n = (2x)^2 + (2y+1)^2 = 4q+1

But, the given number 1234567 = (4×308641)+3 which is of the form 4q+3, hence it cannot be written as the sum of two squares.

To know more about the even numbers

https://brainly.com/question/2289438

#SPJ4