Question

show that square if any positive integer is either of the form 4q or 4q + 1 for some integer q.

Answer 1

Hey there!

According to Euclid division lemma, There exists two integers q, r for a, b which are positive integers such that ,

$\boxed {a = bq + r, 0 ≤ r < b. }$

So, let's consider a = a, b = 4 .

Now, a = 4x + r.

From the above equation,
0 ≤ r < 4 ,
So, r = { 0 , 1 , 2 , 3 }

When r = 0,
then
a = 4x + 0
Squaring on both sides,
a² = ( 4x)²
a² = 16x²
a² = 4 ( 4x²)
a² = 4q , where q = 4x .

When , r = 1
a = 4x + 1
a² = 16x² + 1 + 8x
a² = 4 ( 4x² + 2x ) + 1
a² = 4q + 1 , where q = 4x² + 2x

When r = 2 ,
a = 4x + 2
a² = 16x² + 4 + 16x
a² = 4 ( 4x² + 4x + 1 )
a² = 4q , where q = 4x² + 4x + 1

When r = 3 ,
a = 4x + 3
a² = 16x² + 9 + 24x
a² = 4( 4x² + 6x + 2 ) + 1
a² = 4q + 1 , where q = 4x² + 6x + 2

Therefore , It is proved that Square of any positive integer is of the form 4q or 4q+ 1 for some integer q.