Question

show that the square of any positive odd integer of form 4m+1,for some integer m

Answer 1

Answer:

Step-by-step explanation:

kk

Answer 2

Answer:

Let n be an odd integer.

Then n = 2k+1 for some integer k.

So

n² = ( 2k + 1 )² = 4k² + 4k + 1 = 4 ( k² + k ) + 1 = 4m + 1,

where m = k² + k

That is, n² = 4m + 1.

That is, the square of any odd integer has the form 4m + 1.