Either: a ≡ 0 (mod 2) ... so a^2 ≡ 0*0 (mod 2) ≡ 0 (mod2)
Or: a ≡ 1 (mod 2) ... so a^2 ≡ 1*1 (mod 2) ≡ 1 (mod 2)
So a^2≡a (mod 2) for every integer a
Or you could put it more mundanely:
Integers are either even (0 mod 2) or odd (1 mod 2)
even^2 is even (0 mod 2)
odd^2 is odd (1 mod 2)
So a^2≡a (mod 2) either way
You need to show that 2 divides a^2 - a for every integer a.
Observe a^2 - a = a(a-1). If a is even we're done because then a(a-1) is divisible by 2.
If a is odd then a-1 is even so a(a-1) is divisible by 2 as well.
In any case 2 divides a(a-1) so that a(a-1) â¡ 0 mod 2, i.e a^2 â¡ a mod 2, as desired.
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Either: a ≡ 0 (mod 2) ... so a^2 ≡ 0*0 (mod 2) ≡ 0 (mod2)
Or: a ≡ 1 (mod 2) ... so a^2 ≡ 1*1 (mod 2) ≡ 1 (mod 2)
So a^2≡a (mod 2) for every integer a
Or you could put it more mundanely:
Integers are either even (0 mod 2) or odd (1 mod 2)
even^2 is even (0 mod 2)
odd^2 is odd (1 mod 2)
So a^2≡a (mod 2) either way
You need to show that 2 divides a^2 - a for every integer a.
Observe a^2 - a = a(a-1). If a is even we're done because then a(a-1) is divisible by 2.
If a is odd then a-1 is even so a(a-1) is divisible by 2 as well.
In any case 2 divides a(a-1) so that a(a-1) â¡ 0 mod 2, i.e a^2 â¡ a mod 2, as desired.