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Home » 9. Prove that the square of any positive integer is of the form 5q or 5q + 1, 5q + 4 for some integer q.
9. Prove that the square of any positive integer is of the form 5q or 5q + 1, 5q + 4 for some integer q.
Class 10 | Maths | RD Sharma | Real Numbers
9. Prove that the square of any positive integer is of the form 5q or 5q + 1, 5q + 4 for some integer q.

Solution:

Let ‘a’ be any positive integer.

Then,

According to Euclid’s division lemma,

a=bq+r

According to the question, when b = 5.

    \[a=5k+r,n<r<5\]

When r = 0, we get, a = 5k 

    \[{{a}^{2}}=25{{k}^{2}}=5\left( 5{{k}^{2}} \right)=5q,whereq=5{{k}^{2}}\]

When r = 1, we get, a = 5k + 1

    \[\to {{a}^{2}}={{\left( 5k+1 \right)}^{2}}=25{{k}^{2}}+1+10k=5\left( 5k+2 \right)+1=5q+1\]

    \[q=k\left( 5k+2 \right)\]

When r = 2, we get, a = 5k + 2

    \[\to {{a}^{2}}={{\left( 5k+2 \right)}^{2}}=25{{k}^{2}}+4+20k=5\left( 5{{k}^{2}}+4k \right)+4=4q+\]

    \[4=5{{k}^{2}}+4k\]

When r = 3, we get, a = 5k + 3

    \[\to {{a}^{2}}={{\left( 5k+3 \right)}^{2}}=25{{k}^{2}}+9+30k=5\left( 5{{k}^{2}}+6k+1 \right)+4\]

= 5q + 4, where q = 5k2 + 6k + 1

When r = 4, we get, a = 5k + 4

a2 = (5k + 4)2 = 25k2 + 16 + 40k = 5(5k2 + 8k + 3) + 1

    \[\to {{a}^{2}}={{\left( 5k+4 \right)}^{2}}=25{{k}^{2}}+16+40k=5\left( 5{{k}^{2}}+8k+3 \right)+1\]

= 5q + 1, where q = 5k2 + 8k + 3

Therefore, the square of any positive integer is of the form 5q or, 5q + 1 or 5q + 4 for some integer q.

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