Some basic proof by cases problems

Question

I'm stuck on these basic proof problems. Could you share your solutions with me?

a)Prove 3|(2n^2+1) if and only if 3∤n for n∈Z

b)Let a∈R. Prove that a^2≤1 if and only if -1≤a≤1

Answer 1

Here's part A:

Proves -------> If 3 divides 2N^2+1 then 3 does NOT divide N.

Proof by contradiction:

Suppose 3 Divides N. Then N = 3k for some integer k.

So 2N^2+1 = 2(3k)^2+1 = 18k^2 + 1

Then 3 does not divide 2N^2+1, which is a contradiction.

Proves:

<-------

If 3 does not divide N then 3 divides 2n^2+1.

Since 3 does not divide N, then N = 3k+1 or N = 3k+2

If N = 3k+1 then

2N^2+1 = 2(3k+1)^2+1 = 2(9k^2+6k+1)+1

= 18k^2 + 12k + 3

= 3(6k^2+2k+1)

so 3 divides 2 N^2+1

If n = 3k+2 then

2N^2+1 = 2(3k+2)^2 + 1 = 2(9k^2+12k+4)+1

= 18k^2+24k+9

= 3 (6k^2+8k+3)

so 3 divides 2N^2+1

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Proves

----->

if a^2<=1 then -1 <= a <= 1

Suppose a<-1 or a>1.

If a < -1 Then a^2>1 which is a contradiction.

If a>1 then a^2 > 1 which is a contradiction.

Therefore -1 <= a <= 1

Proves

<------

if -1 <= a <= 1 then a^2<=1

Suppose a^2>1

Then a^2-1 > 0

(a+1)(a-1)>0

Then either a+1>0 and a-1>0 OR a+1<0 and a-1<0

So then either a>-1 and a>1 or a<-1 and a<1

Therefore a>1 OR a<-1 which is a contradiction