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Q) Show that the square of any odd positive integer is of the type 6Q or 6Q+ 3 or 6Q + 5 where Q is some integer.

I am able to find out the 6Q but not 6Q + 3 and 6Q + 5

this is how I did 6Q

let b = 6 so according to the Euclid Division Lemma

a = bq + r

                                                   where 0 ≤ r < b
                                               so, 
                                                   r = 0,1,2,3,4,5

Case - 1

r = 0

a = 6q + 0
Squaring both the sides
(a)² = (6q)²
(a)² = 36q²
(a)² = 6(6q)²                        [where let, Q = 6q²]
(a)² = 6Q

Hence proved

Please like this also prove 6Q + 3 and 6Q + 5

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  • $\begingroup$ Don't you mean $6Q+1, 6Q+3, 6Q+5$?, (because for instance, $17^2=6\cdot48 +1)$. In this case, this question is trivial, as an odd integer squared is odd, and all odd numbers are of the form $6k+1, 6k+3$ or $6k+5, k\in \Bbb Z$. $\endgroup$ Jun 3, 2018 at 13:05
  • $\begingroup$ Welcome to MSE. Please, use MathJax to format your question: math.meta.stackexchange.com/questions/5020/… $\endgroup$
    – Green.H
    Jun 3, 2018 at 13:06
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    $\begingroup$ Welcome to stackexchange. Hint: the way to do these problems is to consider all the possible values an integer might have modulo $6$. What happens if you square $6k+r$ for $r = 0,1,2,3,4,5$? Hint for another method: can you see why the square of an even integer is divisible by$4$, and how that helps? $\endgroup$ Jun 3, 2018 at 13:07
  • $\begingroup$ @RhysHughes 6q + 5 is, however, impossible. $\endgroup$
    – fleablood
    Jun 3, 2018 at 16:14

3 Answers 3

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$(2n + 1)^2 = 4n^2 + 4n + 1$.

Now let $n = 3k +i$ where $i = 0,1,-1$.

$(2n+1)^2 = 4(9k^2 + 6ki +i^2) + 4(3k + i) + 1 =$

$6(6k^2 + 4ki + 2k) + 4i^2 + 4i + 1$

And $4i^2 + 4i + 1 = 1$ if $i = 0$.

$4i^2 + 4i + 1 = 9=6 + 3$ if $i=1$

and $4i^2 + 4i + 1 = $ if $i = -1$

So all odd squares are of then for $6q + 1$ or $6q + 3$ and none are of the form $6q$, $6q + 2$ or $6q + 4$ (which are even and thus impossible) nor of form $6q + 5$.

The statement you wanted to prove is out and out wrong.

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If $a=6q$, then $a$ is even... So that part is wrong.

$a$ odd $\implies a\cong1,3 \text { or } 5 \pmod6$.

Squaring, it is pretty easy to see $a^2\cong1, 9\text { or }25\pmod6$, or $a^2\cong1\text { or }3\pmod6$.

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Every integer $n$ is of the form $6t$, $6t\pm1$, $6t\pm2$, or $6t+3$.

Every odd integer $n$ is of the form $6t\pm1$ or $6t+3$.

Then $n^2$ is of the form $6Q+1$ or $6Q+9$, which reduces to $6Q+3$.

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