## anonymous 4 years ago 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

1. asnaseer

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

2. asnaseer

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

3. asnaseer

therefore:$3m^2(1+n^2)=2^3(251+n^2)$

4. anonymous

Good point asnaseer.

5. asnaseer

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

6. asnaseer

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

7. anonymous

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

8. asnaseer

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.

9. anonymous

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

10. asnaseer

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. :)

11. anonymous

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

12. anonymous

Yes 112 is the answer :)

13. asnaseer

you win gogind :D

14. anonymous

win by brute force, sounds like cheating :D

15. anonymous

haha, prove it now :D

16. asnaseer

need fooood - I'll be back soon :)

17. anonymous

sure :)

18. anonymous

Good thinking gogind.

19. asnaseer

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 $$m^2n=112$$