If k is integer,prove that k squared is congruent to 0 or 1 modulo 4?

Answer 1

there are 2 cases:
k=2m (k even)
k^2=4m^2=0 mod 4

k=2m+1 (k odd)
k^2=4m^2+4m+1 =1 mod 4

Answer 2

Every integer k can be written in the form
k = 4*i + j where i is an integer and 0<= j <= 3

(4*i+j)^2 == 16*i + 8*i*j +j^2 == j^2 mod 4

Remember j is between 0 and 3 inclusive, and
0^2 == 0 mod 4
1^2 == 1 mod 4
2^2 == 0 mod 4
3^2 == 1 mod 4

Answer 3

saying congruent to 0,1 mod 4 means that you want to prove the answer is either divisible by 4 evenly or has a remainder of 1.

for any even integer, k, you can factor out 2 and get k=2n so
k*k=2*n*2*n = 4*n and 4* is evenly divisible by 4 and so any even integer is congruent to 0 mod 4.

for any odd integer, k, we can subtract 2 and get k=2+a
so k*k = (2+a)*(2+a) = 4 + 4a + a*a
the 4 and the 4a when we do modulo 4 get mapped onto 0 leaving only a*a modulo 4. But a is also an odd number and we can go back and do this argument again.
eventually you get to a smaller and smaller a until a = 1 and a*a modulo 4 is simply 1.(proof by induction)

Answer 4

1^2=1*1=1 mod 4
2^2=2*2=0 mod 4
3^2=3*3=1 mod 4
4^2=4*4= 0 mod 4
5^2=5*5=1 mod 4
.....

Answer 5

there are 2 cases
case 1)
k is even so k = 2n so k^2 = 4n^2 so k^2 mod 4 = 0
case 2)
k is odd so k = 2n+1 so k^2 = 4n^2+4n+1 = 4n(n+1) +1
so k^2 mod 4 = 1

hence k^2 mod 4 = 0 or 1

Answer 6

If k is even, then write k as 2x. Squaring this gives 4x^2, which is congruent to 0, mod 4.

If k is odd, then write k as (2x+1). Squaring this gives 4x^2 + 4x +1, which is congruent to 1, mod 4.

Hope this is starting to make sense... :)