Show that the squares of the elements in Z4 are only 0 and 1

Show that the squares of the elements in Z_4 are only [0] and [1]. (b) If a, b are arbitrary integers, show that [a^2] + [b^2] notequalto [3] in Z_4. (c) Show that the equation a^2 + b^2 = 3c^2 has no solutions in non-zero integers a, b, c.

Solution

(a) We need to show that squares of all the elements of Z4 are only [0] and [1]

Let us check this all the elements of Z4. The elements of Z4 are 0, 1, 2 and 3.

0² = [0]

1² = [1]

2² = 4 (mod 4) = [0]

3² = 9 (mod4) = [1]

Therefore, square of all the elements in Z4 is either [0] or [1].

(b) If a and b are arbitrary integers, then we need to show that sum of squares of any two elements of Z4 cannot be 3.

As we have already seen in the previous part that square of any element in Z4 will result in either [0] or [1]. Therefore, sum of any two squares in Z4 can maximum be equal to 2. Hence, sum of squares of any two elements in Z4 cannot be equal to [3]. Therefore, the given statement is true.

(c) Let us consider the two cases of c^2:

If c^2 = [1], then a^2+b^2 = 3 (which is not possible, see part (b))

If c^2 = [0], then a^2+b^2 = 0. (For this to be possible, at least one of a, b and c needs to be 0. Which is a contradiction to the given question)

Hence, the given equation doesn\'t have non solutions.