Abstract Algebra Help?

Question

If a an integer, prove that a^2 is not congruent to 2 modulo 4 or to 3 modulo 4.

also I need help with this proof:

prove that every odd integer is congruent to 1 modulo 4 or to 3 modulo 4.

Answer 1

The a^2 case. Write a as a=4 k + b, for some k and b=0,1,2 or 3.
Now square left and right hand side. What do you conclude for the three terms, especially the last one b^2?

The odd integer: write a as a= 4 k + b . What values can b take on?

Answer 2

By the definition of congruence,
a^2 = 2(mod 4) implies that a^2 - 2 = 4n for some integer n. Now let a be even, then a = 2k for k an integer. Then from the above equation:
4k^2-2 = 4n or k^2-1/2 = n. Now if k is an integer, then so is k^2, thus n cannot be an integer, so the original assumption is false! If instead a is odd, then a = 2k+1 for integer k. Plug this into the congruence equation to get
4k^2+4k+1-2 = 4n or k^2+k-1/2 = n. Again, since k is an integer the left hand side of this equation is not an integer, therefore n is not an integer, and we have another contradiction.

Answer 3

1st answer contained a typo which this corrects:

if a=2k+1
==>a^2=4k^2+4k+1
==>a^2=4(k^2+k)+1
if a=2k
==> a^2=4k^2
there is no other option for a to be 2k+2 or anything else cuz
2k+2=2(k+1) and can be explain by 2k
so a^2 is congruent to 1 or 0 modulo 4
------------------------------
every odd integer=2k+1
k=2n or 2n+1
so
odd integer=2(2n)+1=4n+1
or
odd integer=2(2n+1)+1=4n+3

Answer 4

if a=2k+1
==>a^2=4k^2+4k+1
==>a^2=4(k^2+k)+1
if a=2k
==> a^2=4k^2
there is no other option for a to be 2k+2 or anything else cuz
2k+2=2(k+1) and can be explain by 2k
so a^2 is congruent to 1 or 2 modulo 4
-------------------------------------------------------
every odd integer=2k+1
k=2n or 2n+1
so
odd integer=2(2n)+1=4n+1
or
odd integer=2(2n+1)+1=4n+3
sorry for my English.