Q. 93.9( 23 Votes )

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

Answer :

Since any positive integer n is of the form 5p or 5p + 1, or 5p + 2 or 5m + 3 or 5p +4.

When n = 5p, then

n^{2} = (5p)^{2} = 25p^{2}⇒ 5 (5p^{2}) = 5a, where a =5p^{2}

When n = 5p + 1,

then n^{2} = (5p + 1)^{2} = 25p^{2} + 10p + 1

⇒ 5p(5p + 2) + 1 ⇒ 5a + 1 Where a = p(5p + 2)

When n = 5p + 2,

then n^{2} = (5p + 2)^{2}⇒ 25p^{2} + 20p + 4 ⇒ 5p(5p + 4) + 4

⇒ 5a + 4 where a = p(5p + 4)

When n = 5p + 3, then n^{2} = 25p^{2} + 30p + 9

⇒ 5(5p^{2} + 6p + 1) + 4 ⇒ 5a + 4 where a = 5p^{2} + 6m + 1

When n = 5p + 4, then n^{2} = (5p + 4)^{2} = 25p^{2} + 40p + 16 ⇒ 5(5p^{2} + 8m + 3) + 1 = 5a + 1 where a = 5p^{2} + 8m + 3

Therefore from above results we got that n2 is of the form 5q or, 5q + 1 or, 5q + 4.

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