Function Equation with Even/Odd Cases

Question

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

Accepted Answer

10. Case 1: $m$ is odd $\implies m+1$ is even. $f(m) = m+3$, $f(m+1) = (m+1)(m+2)$. $8(m+1)(m+2) - (m+3) = 2 \implies 8m^2 + 23m + 11 = 0$, no positive integer solution. Case 2: $m$ is even $\implies m+1$ is odd. $f(m) = m(m+1)$, $f(m+1) = (m+1)+3 = m+4$. $8(m+4) - m(m+1) = 2 \implies 8m + 32 - m^2 - m = 2 \implies m^2 - 7m - 30 = 0$. $(m-10)(m+3) = 0 \implies m = 10$.

Answer

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Function Equation with Even/Odd Cases

Explanation

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