Rename $\{a\}=\alpha$ $[a]=m$ $[b]=p$ Now we rewrite the equality in: $r ( \alpha + m) + s \frac 1{\alpha} + t \frac{\alpha}{1-p\alpha} =0$ Mutiplying per $\alpha(1-p\alpha)$ we get: $tp\alpha^3+(tmp-t-r)\alpha^2 + (ps-tm)\alpha - s=0$ Now if $(m+0.5)^3<n<(m+1)^3$ we have that $p=1$ in fact we have $\alpha>0.5$ and so $1<b<2$. Now we take $t=1$, $r=-1-2m$ and $s=3m^2+m$ and we have: $\alpha^3+(m-1+1+2m)\alpha^2 + (3m^2+m-m)\alpha - 3m^2-m=0$ $\alpha^3+3m\alpha^2+3m^2\alpha-3m^2-m=0$ $(\alpha+m)^3-3m^2-m-m^3=0$ $n=m^3+3m^2+m$ and so because we have $(m+0.5)^3<n<(m+1)^3$ if $m>0$ we can say that for $n=m^3+3m^2+m$ with $m \in N$ we can find integers $(r,s,t)=(-1-2m,3m^2+m,1)$ such that $ra+sb+tc=0$ is it ok? btw... nice problem!

Simo_the_Wolf wrote: Mutiplying per $\alpha(1-p\alpha)$ we get: $tp\alpha^3+(tmp-t-r)\alpha^2 + (ps-tm)\alpha - s=0$ is it ok? I think you've made a typo.Am I right?

Since $ c $ is the "nastiest" of the variables we want to artificially make it nice. A good method to do so would be to force $ [b] = 1 $. Let $ \{\} $ denote the fractional part of a real number. Since $ b = \frac{1}{\{a\}} $ we want $ \{a\} > \frac{1}{2} $ so for now assume this to be the case. Now, let $ \alpha = \{a\} $ and $ m = [a] $. We have that $ a = m + \alpha $ and $ b = \frac{1}{\alpha} $ and $ c = \frac{\alpha}{1 - [b]\alpha} = \frac{\alpha}{1 - \alpha} $. So we want $ r(m + \alpha) + \frac{s}{\alpha} + \frac{t\alpha}{1 - \alpha} = 0 $. Multiplying both sides by $ \alpha(1 - \alpha) $ we get that $ -r\alpha^3 + (r - rm + t)\alpha^2 + (mr - s)\alpha + s = 0 $. Since we want to figure out what $ n $ should be in terms of $ m $ we should choose $ r, s, t $ so that we can find a $ (m + \alpha)^3 $ in our equation. So we want to solve the simultaneous set of equations in terms of $ m $: \[ -r = -1 \] \[ r - rm + t = -3m \] \[ mr - s = -3m^2 \] Therefore we find that $ r = 1, s = 3m^2 + m, t = -2m - 1. $ Plugging these in our equation becomes $ m^3 + 3m^2 + m - (m + \alpha)^3 = 0 \Longrightarrow n = m^3 + 3m^2 + m $. Now we need to check that $ \alpha > \frac{1}{2} $ like we assumed but this is clear since $ m^3 + 3m^2 + m > \left(m + \frac{1}{2}\right)^3 $ for all positive integers $ m $. Therefore for every $ n $ of the form $ m^3 + 3m^2 + m $ for some positive integer $ m $, letting $ r = 1, s = 3m^2 + m, t = -2m - 1 $ yields a desired triple $ (a, b, c) $. Therefore we have found an infinite number of solutions and so we are done.

My solution is almost the same, but motivated in another way. We see that it suffices to show that $xa + yb + zc=0$, for some $(x,y,z) \in \mathbb{Q}- \{(0,0,0)\}$, or equivalently \[\frac{xa}{bc} + \frac{y}{c} + \frac{z}{b} = 0 \qquad (1)\].The integer part bothers us, so we let $n=k^3+i$ for $k$ and $0 < i < 3k^2+3k+1 $ to be constructed. We have easily $[n]=k$, and \[ b = \frac{1}{a-[a]} = \frac{1}{\sqrt[3]{n} - k} = \frac{\sqrt[3]{n^2}+k \sqrt[3]{n} + k^2}{i} \]. Now let $M = [b]$, to be easier to deal with for now. We have $(1)$ becomes after expanding, rearranging and multiplying by $i$; \[ \left[ y-iMkx \right]\sqrt[3]{n^2} + \left[ x(i+iMk)+yk+z \right] \sqrt[3]{n} + \left[ (k^2-iMk)y - kz \right] = 0\qquad (2) \].The following is a known fact; So in order to have a solution for problem, It is necessary to find solutions in $x,y,z$ in rational number set, for the equations; \[\begin{cases} \begin{array}{lcl} -iMkx + y & = & 0 \\ x(i+iMk)+yk+z & = & 0 \\ (k^2-iMk)y \quad - kz & = & 0 \\ \end{array} \end{cases} \] From linear algebra we know in order to get a non-trivial solution (in fact infinitely many of them) for the system of equations above, we must have \[ \begin{vmatrix} i+iMk \quad & k & 1 \\ -iMk \quad & 1 & 0 \\ 0 \quad & k^2-iMk& -k \end{vmatrix} = 0 \] After evaluating the above determinant, we get equivalent equation after dividing by $-i$ ; \[k + 3Mk^2 - iM^2 =0 \], It is now plausible to think of $i=3k+k$ and $M = 1$. In fact that can be achieved for infinitely many $n$'s, where we note that if we set $n=k^3 + 3k^2 + k$, we have \[ \frac{\sqrt[3]{n^2}+k \sqrt[3]{n} + k^2}{i} = \frac{\sqrt[3]{(k^3+3k^2+k)^2}+k \sqrt[3]{k^3+3k^2+k} + k^2}{3k^2+k} \rightarrow 1 \] as $k \rightarrow \infty $, where the limit clearly approaches 1 from above. So for large enough $k$, we have $M=1$, so we get the following family of infinitely many non-cubic integers, for which the condition is satisfied which is \[ \{k^3 + 3k^2 + k : k \geq N \} \], where $N$ is sufficiently large integer, It is clear one can let $N=1$.

One of the failed paths that I walked down produced this problem: Let $n=m^3+k$. Show that if $k\mid 3m^2$, then $a, b, c$ are never linearly dependent.

bobthesmartypants wrote: One of the failed paths that I walked down produced this problem: Let $n=m^3+k$. Show that if $k\mid 3m^2$, then $a, b, c$ are never linearly dependent. I hit the same problem during my solution. I just eventually set $\lfloor b \rfloor = \frac{3m^2+r}{k}$ and bashed from there, which gave the interesting condition that $9m^4+4mk$ is a square (which gives $n=m^3+3m^2+m$, but is kind of interesting in its own right).

Let $n=m^3+3m^2+m$. Observe that for all positive $m$, $m^3+3m^2+m>(m+0.5)^3$ so we have \[b=\frac{1}{\sqrt[3]{m^3+3m^2+m}-m}<\frac{1}{m+0.5-m}<2\]which implies that $\lfloor b\rfloor=1$. Hence, \[c=\frac{1}{b-1}=\frac{\sqrt[3]{m^3+3m^2+m}-m}{m+1-\sqrt[3]{m^3+3m^2+m}}\]Let $(r,s,t)=(1,3m^2+m,-2m-1)$. We have \begin{align*} ra+sb+tc &= a + \frac{3m^2+m}{a-m} - \frac{(2m+1)(a-m)}{m+1-a} \\ &= a + \frac{a^3-m^3}{a-m} - \frac{((m+1)^3-a^3)(a-m)}{m+1-a} \\ &= a + a^2 + am + m^2 - ((m+1)^2+a^2+a(m+1))(a-m) \\ &= a^2+am+m^2+a - (m^2+a^2+am)(a-m) - (2m+1+a)(a-m) \\ &= a^2+am+m^2+a - (a^3-m^3) - (am+a+a^2-2m^2-m) \\ &= m^2 - (3m^2+m) + (2m^2+m) \\ &= 0 \end{align*}As desired.

I claim that all numbers of the form $n = k^3+3k^2+k$ satisfy the conditions. Observe that $n > \left(m+\frac 12\right)^3$. Now write \begin{align*} b &= \frac 1{\sqrt[3]{n} - m} = \frac{\sqrt[3]{n^2} + k \sqrt[3]{n} + k^2}{3k^2 + k} \\ c &= \frac 1{b-1} = \frac{k^2 + (k+1) \sqrt[3]{n} + \sqrt[3]{n^2}}{2k+1}. \end{align*}Notice that $c = \frac 1{b-1}$ as $\lfloor b \rfloor = 1$ stemming from $n > \left(m+\frac 12\right)^3$. I claim that the triple $(r, s, t) = (1, 3k^2+3, -(2k+1))$ satisfies the conditions. Let $x = \sqrt[3]{n}$, so that $a \equiv a(x)$, $b \equiv b(x)$, $c \equiv c(x)$ are polynomials in $x$. I will show that $r \cdot a(x) + s \cdot b(x) + t \cdot c(x)$ is identically zero. In particular, looking at each term, \begin{align*} a_0 &= (3k^2+k) \left( \frac{k^2}{3k^2+k} \right)- (2k+1) \left(\frac{k^2}{2k+1} \right)= 0 \\ a_1 &= 1 + (3k^2+k) \left(\frac k{3k^2+k}\right) - (2k+1) \left(\frac{k+1}{2k+1}\right) = 0 \\ a_2 &= (3k^2+k)\left(\frac 1{3k^2+k}\right) - (2k+1) \left(\frac 1{2k+1}\right) = 0. \end{align*}Thus $ra+sb+tc = 0$, as required. Remark: The only difficult part of the problem is picking $\lfloor b \rfloor = 1$. This comes from the fact that as long as $n - k^3$ is ``decently large", we ought to have $\lfloor b \rfloor = 1$; by picking this, we leave ourselves with a pretty wide range of $n$ that we could still consider. This also makes the expression for $c$ much cleaner.