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 $8 f(m+1)-f(m)=2$, then $m$ equals

We have a piecewise function $f(n)$ that behaves differently based on whether $n$ is even or odd:

If $n$ is even: $f(n) = n(n+1)$

If $n$ is odd: $f(n) = n+3$

We need to find the positive integer $m$ such that $8f(m+1) - f(m) = 2$.

Key Insight: Since we don't know if $m$ is even or odd we need to check both possibilities.

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If $m$ is odd then $m+1$ must be even.

$f(m+1) = (m+1)(m+2)$ because $m+1$ is even

$f(m) = m+3$ because $m$ is odd

$8f(m+1) - f(m) = 2$

$8(m+1)(m+2) - (m+3) = 2$

$8(m+1)(m+2) = 8(m^2 + 3m + 2) = 8m^2 + 24m + 16$

$8m^2 + 24m + 16 - m - 3 = 2$

$8m^2 + 23m + 13 = 2$

$8m^2 + 23m + 11 = 0$

For a quadratic $ax^2 + bx + c = 0$ we can find integer solutions only if the discriminant $b^2 - 4ac$ is a perfect square.

Discriminant = $(23)^2 - 4(8)(11) = 529 - 352 = 177$

Since $\sqrt{177}$ is not a whole number this case doesn't give us integer values for $m$.

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If $m$ is even then $m+1$ must be odd.

$f(m+1) = (m+1) + 3 = m + 4$ because $m+1$ is odd

$f(m) = m(m+1)$ because $m$ is even

$8f(m+1) - f(m) = 2$

$8(m + 4) - m(m+1) = 2$

$8m + 32 - m^2 - m = 2$

$7m + 32 - m^2 = 2$

$-m^2 + 7m + 30 = 0$

$m^2 - 7m - 30 = 0$

We need two numbers that multiply to $-30$ and add to $-7$.

Those numbers are $-10$ and $3$.

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

Therefore: $m = 10$ or $m = -3$

Since $m$ must be a positive integer we have $m = 10$.

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Therefore $m = 10$