Prove that the square of any positive integer is of the form 5q, 5q + 1, 5q + 4 for some integer q.
Answers
SOLUTION :
Since positive integer n is of the form of 5m or 5m + 1, 5m + 4.
Case : 1
If n = 5m , then
n² = (5m)²
[On squaring both sides]
n² = 25m²
n² = 5 (5m)
n² = 5q (Where q = 5m)
Case : 2
If n = 5m + 1, then
n² = (5m +1)²
[On squaring both sides]
n² = (5m)²+ 10m + 1
[(a+b)² = a² + b² + 2ab]
n² = 25m² + 10m + 1
n² = 5m (5m + 2) + 1
n² = 5q +1 , where q = m (5m + 2)
Case : 3
If n = 5m + 2, then
n² = (5m + 2)²
[On squaring both sides]
n² = (5m)² + 20m + 4
[(a+b)² = a² + b² + 2ab]
n² = 25m² + 20m + 4
n² = 5m (5m + 4) + 4
n² = 5q + 4 (where q = m (5m + 4))
Case : 4
If n = 5m + 4, then
n²= (5m + 4)²
[On squaring both sides]
n²= (5m)² + 40m + 4²
[(a+b)² = a² + b² + 2ab]
n² = 25m² + 40m + 16
n² = 5 (5m² + 8m + 3) + 1
n² = 5q + 1 , where q = 5m² + 8m + 3 )
Hence ,it is proved that the square of any positive integer is of the form 5q or 5q + 1, 5q + 4 for some integer q.
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Hey there !
Solution:
According to Euclid's Lemma, any number is of the form : a = bq + r
According to the question, Let b = 5. So r = 0, 1, 2, 3 and 4.
Case 1:
a = 5q + 0
Squaring on both sides, we get,
=> a² = ( 5q )² = 25q²
This can be written as a² = 5 ( 5q² ) or 5 ( x ). Here x = 5q².
So it is of the form 5 ( x )
Case 2:
a = 5q + 1
Squaring on both sides we get,
=> a² = ( 5q + 1 )²
=> a² = ( 25q² + 10q + 1 )
=> a² = 5 ( 5q² + 2q ) + 1
This is of the form, 5 ( x ) + 1 where x = ( 5q² + 2q ).
Case 3:
a = 5q + 2
Squaring on both sides we get,
=> a² = ( 5q + 2 )²
=> a² = ( 25q² + 20q + 4 )
=> a² = 5 ( 5q² + 4q ) + 4 which is of the form 5 ( x ) + 4,
where x = ( 5q² + 4q ).
Case 4:
a = 5q + 3
Squaring on both sides, we get,
=> a² = ( 5q + 3 )²
=> a² = ( 25q² + 30q + 9 )
=> a² = ( 25q² + 30q + 5 + 4 )
=> a² = 5 ( 5q² + 6q + 1 ) + 4 which is of the form 5 ( x ) + 4,
where x = ( 5q² + 6q + 1 )
Case 5:
a = 5q + 4
Squaring on both sides, we get,
=> a² = ( 5q + 4 )²
=> a² = ( 25q² + 40q + 16 )
=> a² = ( 25q² + 40q + 15 + 1 )
=> a² = 5 ( 5q² + 8q + 3 ) + 1, which is of the form 5 ( x ) + 1,
where x = ( 5q² + 8q + 3 ).
Hence in all the cases, the form we get for a square of positive number is either 5x, 5x + 1 or 5x + 4.
Hence the square of any positive integer will always be of the form:
5x, 5x + 1 or 5x + 4.
Hence Proved !
Hope my answer helped !