Question

Fool's problem of the day,

If \( m \) and \(n\) are positive integers that satisfy \(3m^2 − 8n^2 + 3m^2n^2 = 2008\), then find the product \(m^2n\).

Genre: algebra-precalculus
Rating: Easy

Answer 1

I have a feeling that we should make use of the fact that \(2008=2^3*251\)

Answer 2

I don't know if this helps, but we can rewrite this as:\[3m^2(1+n^2)-2^3n^2=2^3*251\]

Answer 3

therefore:\[3m^2(1+n^2)=2^3(251+n^2)\]

Answer 4

Good point asnaseer.

Answer 5

therefore:\[m^2n=\frac{2^3n(251+n^2)}{3(1+n^2)}\]

Answer 6

this must be an integer, therefore \(3(1+n^2)\) must divide evenly into \(2^3n(251+n^2)\)

Answer 7

You are in the right track, divisibility the way :)

Answer 8

since \(2^3\) is not divisible by 3, then this implies \(n(251+n^2)\) must be divisible by 3. therefore either n is divisible by 3 and/or \(251+n^2\) is divisible by 3.

Answer 9

is it 112? (n = 7), makes (251+n^2) =300 which is nice and makes (1+n^2)=50 also nice :D

Answer 10

I'm trying to /prove/ the result. I am sure we can use brute force to find /a/ solution, but that doesn't prove it's the /only/ solution. :)

Answer 11

ah, nvm them, I don't know how to prove it :D

Answer 12

Yes 112 is the answer :)

Answer 13

you win gogind :D

Answer 14

win by brute force, sounds like cheating :D

Answer 15

haha, prove it now :D

Answer 16

need fooood - I'll be back soon :)

Answer 17

sure :)

Answer 18

Good thinking gogind.

Answer 19

OK - the furthest I can get with this is as follows:\[\begin{align}
3m^2(1+n^2)-8n^2&=2008\tag{a}\\
m^2n&=\frac{8n(251+n^2)}{3(1+n^2)}\tag{b}
\end{align}\]equation (b) implies \(n(251+n^2)\) must be a multiple of 3.
1. lets try n being a multiple of 3.\[\begin{align}
n&=3a\\
3m^2(1+9a^2)-72a^2&=2008\qquad\text{(substituting n=3a into equation(a))}\\
\therefore 3(m^2(1+9a^2)-24a^2)&=2008\\
\qquad&\text{reject since 2008 is not a multiple of 3}
\end{align}\]
2. lets try \(251+n^2\) being a multiple of 3.\[\begin{align}
251+n^2&=3a\\
\therefore n^2&=3a-251\tag{c}
\end{align}\]
Now we can substitute equation (c) into equation (a) to get:\[\begin{align}
3m^2(1+3a-251)-8(3a-251)&=2008\\
3m^2(3a-250)-24a+\cancel{2008}&=\cancel{2008}\\
3m^2(3a-250)&=24a\\
\therefore m^2&=\frac{8a}{3a-250}\tag{d}
\end{align}\]
Equation (d) implies \(a\ge84\). It also shows \(m^2\rightarrow\frac{8}{3}\) as \(a\rightarrow\infty\).

From here I could only use trial and error to determine that the only solution to (c) and (d) is a=100, which gives m=4, n=7.

Therefore \(m^2n=112\)