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Show that square of any positive integer is of the for
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Answers
Step-by-step explanation :-
Euclid's division Lemma any natural number can be written as: .
where r = 0, 1, 2,. and q is the quotient.
∵ Thus any number is in the form of 3q , 3q+1 or 3q+2.
→ Case I: if n =3q
⇒n = 3q = 3(q) is divisible by 3,
⇒ n + 2 = 3q + 2 is not divisible by 3.
⇒ n + 4 = 3q + 4 = 3(q + 1) + 1 is not divisible by 3.
→ Case II: if n =3q + 1
⇒ n = 3q + 1 is not divisible by 3.
⇒ n + 2 = 3q + 1 + 2 = 3q + 3 = 3(q + 1) is divisible by 3.
⇒ n + 4 = 3q + 1 + 4 = 3q + 5 = 3(q + 1) + 2 is not divisible by 3.
→ Case III: if n = 3q + 2
⇒ n =3q + 2 is not divisible by 3.
⇒ n + 2 = 3q + 2 + 2 = 3q + 4 = 3(q + 1) + 1 is not divisible by 3.
⇒ n + 4 = 3q + 2 + 4 = 3q + 6 = 3(q + 2) is divisible by 3.
Thus one and only one out of n , n+2, n+4 is divisible by 3.
Hence, it is solved