Question

prove that the square of a positive integer is of the form 6q+5 then it is of the form 3q+2 for some integer q but not conversely

Answer 1

Let, n = 6q + 5, when q is a positive integer
We know that any positive integer is of the form 3k, or 3k + 1, or 3k + 2
∴ q = 3k or 3k + 1, or 3k + 2
If q = 3k, then
n = 6q + 5
= 6(3k) + 5
= 18k + 5
= 18k + 3 + 2
= 3(6k + 1) + 2
= 3m + 2, where m is some integer
If q = 3k + 1, then
n = 6q + 5
= 6(3k + 1) + 5
= 18k + 6 + 5
= 18k + 11
= 3(6k + 3) + 2
= 3m + 2, where m is some integer
If q = 3k + 2, then
n = 6q + 5
= 6(3k + 2) + 5
= 18k + 12 + 5
= 18k + 17
= 3(6k + 5) + 2
= 3m + 2, where m is some integer
Hence, if a positive integer is of the form 6q + 5, then it is of the form 3q + 2 for some integer q.

Answer 2

Heya !!

Here's your answer.. ⬇⬇
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➡ To Prove :- The square of a positive integer is of the form 6q+5 then it is of the form 3q+2 for some integer q but not conversely.

➡ Proof :- Let n = 6q + 5

We Know that all positive integers are in form of 3m, 3m + 1, and 3m + 2.

Given that we have to take Square of those positive integers.

=> ( 3m )^2 = 9m^2
=> ( 3m + 1 )^2 = ( 9m^2 + 6m + 1 )
=> ( 3m + 2 )^2 = ( 9m^2 + 12m + 4 )

> Taking q = 9m^2

n = 6 ( 9m^2 ) + 5
n = 54m^2 + 3 + 2
Taking 3 as common..,
n = 3 ( 18m^2 + 1 ) + 2
Let ( 18m^2 + 1 ) = k
n = 3k + 2

> Taking q = ( 9m^2 + 6m + 1 )

n = 6 ( 9m^2 + 6m + 1 ) + 5
n = 54m^2 + 36m + 6 + 3 + 2
n = 3 ( 18m^2 + 12m + 2 + 1 ) + 2
Taking ( 18m^2 + 12m + 2 + 1 ) = k
n = 3k + 2

> Taking q = (9m^2 + 12m + 4)

n = 6 ( 9m^2 + 12m + 4 ) + 5
n = 54m^2 + 72m + 24 + 3 + 2
n = 3 ( 18m^2 + 26m + 8 + 1 ) + 2
Taking ( 18m^2 + 26m + 8 + 1 ) = k
n = 3k + 2

Hence, proved.
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Hope it helps..

Thanks :))