prove that the square of any positive integer is of the form 5q, 5q + 1, 5q + 4 for some integer q
Answers
a=bq+r
a=5q+r as b=5
0≤r<5
The possible values of r are 0,1,2,3 and 4
a=1,2,3,4.....
Square of any positive integer:-
(1)²=1=5(0)+1 (r=1)
(2)²=4=5(0)+4 (r=4)
(3)²=9=5(1)+4 (r=4)
(4)²=16=5(3)+1 (r=1)
(5)²=25=5(5)+0 (r=0)
Therefore,the square of any positive integer is of the form 5q,5q+1 or 5q+4.
Hence proved.
Answer:
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.
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
Hence, the square of any positive integer is of the form 5q or 5q + 1 or 51 + 4.
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