use euclid's division Lemma to show that the square of any positive integer cannot be of the form 5 M +2 or 5m+3 for some integer m
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Use euclid's division Lemma to show that the square of any positive integer cannot be of the form 5 M +2 or 5m+3 for some integer m
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Let x be any integer (positive) .
b = 5
By E.D.L we divide it
a = 5m +r where m > 0
Or
π= 1,2,3,4,
a = 5m , a = 5m+1 ,a = 5m+2, a= 5m+3, 5m+4
According to the theorem if the square of any positive integer is divided by then that number is also divided by 2
But the square is not equal to 5m+2,5m+3 proved
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Step-by-step explanation:
Let ‘a’ be the any positive integer .
And, b = 5 .
→ Using Euclid's division lemma :-
==> a = bq + r ; 0 ≤ r < b .
==> 0 ≤ r < 5 .
•°• Possible values of r = 0, 1, 2, 3, 4 .
→ Taking r = 0 .
Then, a = bq + r .
==> a = 5q + 0 .
==> a = ( 5q )² .
==> a = 5( 5q² ) .
•°• a = 5m . [ Where m = 5q² ] .
→ Taking r = 1 .
==> a = 5q + 1 .
==> a = ( 5q + 1 )² .
==> a = 25q² + 10q + 1 .
==> a = 5( 5q² + 2q ) + 1 .
•°• a = 5m + 1 . [ Where m = 5q² + 2q ] .
→ Taking r = 2 .
==> a = 5q + 2 .
==> a = ( 5q + 2 )² .
==> a = 25q² + 20q + 4 .
==> a = 5( 5q² + 4q ) + 4 .
•°• a = 5m + 4 . [ Where m = 5q² + 4q ] .
→ Taking r = 3 .
==> a = 5q + 3 .
==> a = ( 5q + 3 )² .
==> a = 25q² + 30q + 9 .
==> a = 25q² + 30q + 5 + 4 .
==> a = 5( 5q² + 6q + 1 ) + 4 .
•°• a = 5m + 4 . [ Where m = 5q² + 6q + 1 ] .
→ Taking r = 4 .
==> a = 5q + 4 .
==> a = ( 5q + 4 )² .
==> a = 25q² + 40q + 16 .
==> a = 25q² + 40q + 15 + 1 .
==> a = 5( 5q² + 8q + 3 ) + 1 .
∴ a = 5m + 1 . [ Where m = 5q² + 8q + 3 ] .
→ Therefore, square of any positive integer in cannot be of the form 5m + 2 or 5m + 3 .
Hence, it is proved .