Question

If $m,n\in N$ such that $3m^2+m=4n^2+n$, then

• A. $(m-n)(3m+3n+1)$ is divisible by $m^2$.
• B. $(m-n)(3m+3n+1)$ is divisible by $n^2$.
• C. $(m-n)(4m+4n+1)$ is divisible by $m^2$.
• D. $(m-n)(4m+4n+1)$ is divisible by $n^2$.

Answer 1

Answer: (B), (C)

The correct options are
B $(m-n)(3m+3n+1)$ is divisible by $n^2$.
C $(m-n)(4m+4n+1)$ is divisible by $m^2$.

$3m^2+m=4n^2+n$

$⟹3m^2+m=3n^2+n^2+n$
$⟹3m^2-3n^2+m-n=n^2$

$⟹3(m^2-n^2)+m-n=n^2$

$⟹3(m-n)(m+n)+m-n=n^2$
$⟹(m-n)(3m+3n+1)=n^2$
$\therefore$ $(m-n)(3m+3n+1)$ is divisible by $n^2$

Option B.

Also, $3m^2+m=4n^2+n$

$⟹4m^2+m-4n^2-n=m^2$

$⟹4(m^2-n^2)+(m-n)=m^2$
$⟹4(m-n)(m+n)+(m-n)=m^2$
$⟹(m-n)(4m+4n+1)=m^2$
$\therefore$ $(m-n)(4m+4n+1)$ is divisible by $m^2$

Option C.

Hence, options B and C are correct.