Show that for any two integers a, b , (a+b)^2 ≡ a^2 + b^2 (mod 2)(adsbygoogle = window.adsbygoogle || []).push({});

I have my solution below i wanted someone to help chekc if i have done anything wrong. Thank You for your help.

The thing that is going on here is that 2x = 0 (mod 2) for any x. If x = ab, then 2ab = 0 (mod 2).

We see that (a+b)^2 = a^2 + 2ab + b^2. Then this equals a^2 + 0 + b^2 (mod 2).

Therfore, the Proof:

By definition, showing (a+b)^2 = a^2 + b^2 (mod 2) is equivalent to showing that

2 divides [(a+b)^2 - (a^2+b^2)]

2 divides [(a^2 + 2ab + b^2 - a^2 - b^2]

2 divides 2ab

To show 2 divides 2ab we need to find an integer k such that 2k = 2ab. Take k = ab. Thus it is proved.

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# Homework Help: Show that for any two integers a, b , (a+b)^2 ≡ a^2 + b^2 (mod 2)

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