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  • Prove that the square of any positive integer is of the form 5q or,5q+1 or,5q+4 for some integer q.

Prove that the square of any positive integer is of the form 5q or,5q+1 or,5q+4 for some integer q.

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Our aim is to prove that the square of any positive integer is of the form 5q or,5q+1 or,5q+4 for some integer q.

Lets first prove the first part, this is, lets prove that the square of any positive integer is of the form 5q or,5q+1 or 51 + 4.

Let us consider a an integer number such that a = 5m + r.

Applying rules of division algorithm we know that r = 0 or 0 < r < 5.

We need then to consider all cases of r. Thus:

r = 0
Lets also consider q to be equal to m^2.

When r = a we can conclude that a = 5m.

a = 5m

a^2 = ( 5m )^2

a^2 = 5 ( 5m^​2 )

a^2 = 5q.

r = 1
Lets also consider q to be equal to 5m^2 + 2m.

a = 5m + 1

a^2 = ( 5m + 1 )^2 a^2 = 25m^2 + 10m + 1\\ a^2 = 5 ( 5m^2 + 2m ) + 1\\ a^2 = 5q + 1.\\

r = 2
Lets also consider q to be equal to 5m^2 + 4m.

a = 5m + 2\\ a^2 = ( 5m + 2 )^2\\ a^2 = 25m^2 + 20m + 4\\ a^2 = 5 ( 5m^2 + 4m ) + 4\\ a^2 = 5q + 4.

r = 3
Lets also consider q to be equal to 5m^2 + 6m + 1.

a = 5m + 3

a^2 = ​( 5m + 3 )^2

a^2 = 25m^2 + 9 + 30m

a^2 = 25m^2 + 30m ​ + 5 + 4

a^2 = 5 ( 5m^2 + 6m + 1 ) + 4

a^2 = 5q + 4.

r = 4
Lets also consider q to be equal to 5m^2 + 8m + 3.

\begin{align*} a = 5m + 4\\ a^2 = (5m + 4)^2\\ a^2 = 25m^2 + 40m +15 + 1\\ a^2 = 5 ( 5m^2 + 8m + 3 ) + 1\\ a^2 = 5q + 1\\ \end{align*}

Hence, the square of any positive integer is of the form 5q or 5q + 1 or 5q + 4.

Posted by

Satyajeet Kumar

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