# Math Help - determining natural numbers

1. ## determining natural numbers

be u,v natural numbers without common divisor and x,y,z natural numbers each smaller or equal to u*v.
now, prove that it is possible to determine x,y,z in such a way that
0 < sqrt(y² - 4xz) =< 2x/(uv).

well, i have been trying to solve this all night but only could do so applying certain constraints (e.g. that 2 is a factor of either u or v). could anybody help me find a general solution to this?
thanks

2. Originally Posted by tyler4real
be u,v natural numbers without common divisor and x,y,z natural numbers each smaller or equal to u*v.
now, prove that it is possible to determine x,y,z in such a way that
0 < sqrt(y² - 4xz) =< 2x/(uv).

well, i have been trying to solve this all night but only could do so applying certain constraints (e.g. that 2 is a factor of either u or v). could anybody help me find a general solution to this?
thanks
Write w=uv. We want to find natural numbers x, y, z, all less than or equal to w, such that $w\leqslant\frac{2x}{\sqrt{y^2-4xz}}$. ----- (*)

I found that the easiest way to do this is to approach the problem backwards. That is, think of some promising values for x, y and z, and see which values of w they work for.

For example, let $n\geqslant2$ and take $x=n(n+1)$, $y=2n+1$ and $z=1$. Then $y^2-4xz=1$ and the equation (*) becomes $w\leqslant2x=2n(n+1)$. The greatest of the numbers x,y and z is x, namely n(n+1). So that choice $(x,y,z) = (n(n+1),2n+1,1)$ will work for all numbers w in the interval $n(n+1)\leqslant w\leqslant2n(n+1)$.

But each of those intervals overlaps with the next one, since $(n+1)(n+2)\leqslant 2n(n+1)$. So taken together, those intervals cover all the natural numbers, except for those that lie before the start of the first interval. That is the interval corresponding to n=2, which covers the numbers $6\leqslant w\leqslant 12$. So we are left with having to deal with w=1,2,3,4 and 5.

For w=4 we can take x=2, y=3 and z=1. But now we can make use of the information that w is of the form uv (where u and v have no common divisor). I think that the question probably also wants us to assume that neither u nor v is equal to 1. If so, then the numbers 1,2,3 and 5 are not of the form uv so we don't have to worry about them anyway, and the proof is complete.

Edit. You may want to require that x, y and z should also be different from 1. In that case, take $(x,y,z) = (n+1,2n+1,n)$, for n=2,3,4,..., and use a similar line of reasoning.