show that the square of any odd integer is of the form 4q+1 for some integer q.

Question

shubhamsrg · Answer

any odd integer is in the form of (2n +1),,where n is some integer
sq of it = 4n^2 + 1 + 4n = 4(n^2 +n) + 1 = 4q + 1
where q=(n^2 +n)
hence proved..
hmmn

kinggeorge · Answer

On a related tangent, the square of any odd integer is also of the form 8k+1 for some integer k. This is a slightly more involved proof (but only slightly).

anonymous · Answer

can any1 give step by step?

kinggeorge · Answer

@shubhamsrg already gave the step by step for the 4q+1 problem. If you want one for the 8k+1 problem, I'll be happy to elaborate.

shubhamsrg · Answer

induction may also be used here as a successful tool..

shubhamsrg · Answer

both for 4k+1 and 8k+1

shubhamsrg · Answer

and for that 8k+1 by my method,,we know n^2 + n = n(n+1) which is always even (ofcorse)
= 2k,for some integer k..
hence it takes the shape of 8k+1
is there any other approach than this @KingGeorge sir ?

kinggeorge · Answer

There are a few ways to approach the 8k+1 problem. Simplest, is modular arithmetic. Next simplest, is the way you just described. More complicated, but still straightforward, is to look at $4n^2+4n+1$ and look at the case where $n$ is even vs. $n$ is odd. There's also induction, as you also mentioned.

shubhamsrg · Answer

yeah i see what you're intending to explain :)

mathslover · Answer

@jyosnika  shubhamsrg has the right answer ...