Question #8531

let m, n be an element of Z. Using Congruence of Integers, prove that if n is logically equivalent to 1 (mod 2) and m is logically equivalent to 3 (mod 4), then n^2 + m is logically equivalent to 0 (mod 4)

Expert's answer

n = 1 (mod 2)

So we can write:

n = 2k+1

where k is an integer.

We have:

n2 = (2k+1)2 = 4k2 + 4k + 1

On the other hand:

m = 3 (mod 4)

so we can write:

m = 4q+3

where q is an integer.

therefore:

n2 + m = 4k2 + 4k + 1 + 4q+3 = 4k2 + 4k + 4 + 4q = 4(k2 + k + 1 + q)

So we have:

n2 + m = 0 (mod 4).

So we can write:

n = 2k+1

where k is an integer.

We have:

n2 = (2k+1)2 = 4k2 + 4k + 1

On the other hand:

m = 3 (mod 4)

so we can write:

m = 4q+3

where q is an integer.

therefore:

n2 + m = 4k2 + 4k + 1 + 4q+3 = 4k2 + 4k + 4 + 4q = 4(k2 + k + 1 + q)

So we have:

n2 + m = 0 (mod 4).

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