Show that a^2~0 or a^2~1 in Mod 4

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SUMMARY

The discussion confirms that for any integer \( a \), \( a^2 \equiv 0 \) or \( a^2 \equiv 1 \) in modulo 4. The solution utilizes the division algorithm, demonstrating that even integers can be expressed as \( a = 2q \), leading to \( a^2 \equiv 0 \mod 4 \). Conversely, for odd integers represented as \( a = 2q + 1 \), the result \( a^2 \equiv 1 \mod 4 \) is established. This method effectively validates the equivalence for both even and odd integers.

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Homework Statement


If a I san integer, show that a^2~0 or a^2~1 in mod 4 (~ represent equivalence)


Homework Equations





The Attempt at a Solution


my ATTEMPT:
I started with the division algorithm..
a = 2q + 1 for all odd numbers
a = 2q + 0 for all even numbers
then I squared the formula for even numbers...
a^2 = 4q^2 + 0... but q is arbitrary so
a^2 ~ 0 mod 4 for even numbers

then I did the same for odd numbers
a^2 = 4q^2 + 4q + 1
and then I said since q was arbitrary again I could say
a^2 = 4q + 1
a^2~1 mod 4 for odd numbers


is this the correct approach?
 
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Yep, clean work.
 

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