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For any positive integer $n$, let $f(n)=n(n+1)$ if n is even, and $f(n)=n+3$ if n is odd.  if $m$ is a positive integer such that $8f(m+1)-f(m)=2$, then $m$ equals _______
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Given that $,f(n) = \left \{\begin{matrix} n(n+1)\;; &  \text {if n is even} \\ (n+3)\;;  & \text {if n is odd} \end{matrix}\right.$

And $,8f (m+1) – f(m) = 2 $

$ \textbf{Case 1:}$ If $`m\text{’}$ is an even, then $`m+1\text{’}$ must be odd.

$ 8f (m+1) – f(m) = 2 $

$ \Rightarrow 8(m+1+3) – m(m+1) = 2 $

$ \Rightarrow 8(m+4) – m^{2} – m = 2 $

$ \Rightarrow 8m + 32 – m^{2} – m = 2 $

$ \Rightarrow -m^{2} +7m +30 = 0 $

$ \Rightarrow m^{2} – 7m - 30 = 0 $

$ \Rightarrow m^{2} – 10m + 3m - 30 = 0 $

$ \Rightarrow m(m-10) + 3(m-10) = 0 $

$ \Rightarrow (m-10) (m+3) = 0 $

$ \Rightarrow m-10=0, m+3=0 $

$ \Rightarrow \boxed{m=10}, \boxed{m=-3} $ [rejected $`m\text{’}$ is an even positive integer.]

$ \Rightarrow \boxed{m=10} $

$ \textbf{Case 2:}$ If $`m\text{’}$ is an odd, then $`(m+1)\text{’}$ must be an even.

$8f (m+1) – f(m) = 2 $

$ \Rightarrow 8 (m+1) (m+1+1) – (m+3) = 2 $

$ \Rightarrow 8 (m+1) (m+2) – m – 3 – 2 = 0 $

$ \Rightarrow 8 ( m^{2} + 2m + m + 2 ) – m – 5 = 0 $

$ \Rightarrow 8m^{2} + 16m + 8m + 16 – m – 5 = 0 $

$ \Rightarrow 8m^{2} + 24m + 16 – m – 5 = 0 $

$ \Rightarrow 8m^{2} + 23m + 11 = 0 $

We know that, factors for, $ax^{2} + bx + c = 0 \Rightarrow x = \frac{-b \pm \sqrt{b^{2} – 4ac}}{2a}$

Here, $m = \frac{-23 \pm \sqrt{(23)^{2} – 4(8)(11)}}{16}$

$\Rightarrow m = \frac{-23 \pm \sqrt{529 – 352}}{16}$

$\Rightarrow m = \frac{-23 \pm \sqrt{177}}{16}$

Here, the value of $`m\text{’}$ is not integer.

So, no integer solution is possible for $`m\text{’}$ in the case.

Therefore$, \boxed{m = 10} $

Correct Answer $:10$
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