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$x$, $y$, and $z$ are three positive integers, with $x > z$.

$(x, 2y, z)$ is an arithmetic progression

$(x, y, z)$ is a geometric progression.

Find the value of $z/x$.

I mapped $(x, 2y, z)$ to $(a, a + d, a + 2d)$ and $(x, y, z)$ to $(a, ar, ar^2)$

My current approach: find $r$ by dividing $ar$ by $a$, and $ar^2$ by $ar$. Set these 2 equations equal to each other. Which works out to be $\ (a+2d) / 2a = (2a + 4d) / (a +d) = r $.

I am stuck here, what should my next step be?

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Gen Tan
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3 Answers3

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With three term progressions, I think it is easier to fix the middle term. Thus, in this case, we write the arithmetic progression as $$(x,2y,z)=(2y-P,2y,2y+P)$$ and the geometric as $$(x,y,z)=\left(\frac yt,y,yt\right)$$

On first glance it appears that we have two equations in the three variables $(y,P,t)$ but we can eliminate one of these. Nothing in the problem changes if we scale both triples (the a.p. and the g.p. remain in progression and the ratio of the outer terms is unchanged). Thus we may assume that $y=1$ so we have $$(2-P,2,2+P)\quad \&\quad \left(\frac 1t,1,t\right)$$

(Note: the value of $P$, though not of $t$, changed after the scaling, but we will continue to use the same variable for simplicity. The scale factor will cancel out in the final ratio).

We now have two equations in two unknowns. So:

$$2-P=\frac 1t\quad \&\quad 2+P=t\implies 2-P=\frac 1{2+P}\implies 4-P^2=1\implies P=-\sqrt 3$$

Note: we are told that $x>z$ which implies that $P<0$.

Thus $$P=-\sqrt 3\quad \& \quad t=2-\sqrt 3$$ and the desired ratio is $$\boxed {\frac {2-\sqrt 3}{2+\sqrt 3}=\left(2-\sqrt 3\right)^2=7-4\sqrt 3}$$

Note: this solution is not compatible with the stated condition that $x,y,z\in \mathbb N$. Indeed, I do not see how that condition can be satisfied. If there were a solution over the natural numbers then the ratio $\frac zx$ would clearly be rational, which my solution is not. So, either I am mistaken or the extra condition means there is no solution.

lulu
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  • $$ z \lt x \implies z/x \lt 1 \ &, x,,y,,z \in\mathbb{N}^{+} $$ – Hazem Orabi Sep 25 '18 at 10:37
  • @HazemOrabi You are correct, thanks. Indeed, $x>z \implies P<0$ despite what I said. I'll edit accordingly. – lulu Sep 25 '18 at 10:39
  • go go lulu ... awaiting 4 your update. take-care. – Hazem Orabi Sep 25 '18 at 10:44
  • @HazemOrabi Changing the sign of $P$ only took a second. Still, though, my method does not yield integers. I don't see how that condition is possible. To stress: unless I am mistaken (always possible), my method yields all possible solutions and it seems fairly clear that no re-scaling of my core solution gets you all integers. Indeed, if all three are integers then the ratio has got to be rational. – lulu Sep 25 '18 at 10:45
  • 100% agree with u. I'm clueless. I'll try something, meanwhile (+1). Thanks. – Hazem Orabi Sep 25 '18 at 10:53
  • @HazemOrabi If my claim regarding the scaling is correct, then I have the only ratio that works. I don't see any problem with the scaling...I just multiply both triples by $\frac 1y$...doesn't change anything, certainly doesn't change the ratio. Anyway, if I have a blunder somewhere, I expect it's there. – lulu Sep 25 '18 at 10:54
  • small typo, which effect nothing: $$ \left(y/t,,y,,\color{red}{y},t\right) $$ – Hazem Orabi Sep 25 '18 at 11:01
  • @HazemOrabi corrected, thanks again. – lulu Sep 25 '18 at 11:05
  • Even without scaling, your answer is correct. $$ \frac{z}{x}=\frac{y,t}{y/t}=t^2 $$ Solving the two equations with respect to $,t,$: $$\begin{align} & y/t=2y-p \implies p=2y-y/t \ & y,t=2y+p=4y-y/t \implies \ & t=4-1/t \implies \color{red}{t=2\pm\sqrt{3}} \end{align}$$ – Hazem Orabi Sep 25 '18 at 11:33
  • @HazemOrabi Sure...the scaling trick just makes the algebra cleaner. – lulu Sep 25 '18 at 11:48
  • @lulu Do you happen to know how I can change $B1$ in the ecm-routine in yafu , version $1.34\ $ ? – Peter Sep 25 '18 at 14:14
  • @Peter Sorry, never used yafu. – lulu Sep 25 '18 at 14:17
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Here is an algebraically a bit more direct way:

  • Since $(x,2y,z)$ is an arithmetic progession, we have: $$2y = \frac{x+z}{2} \Leftrightarrow y = \frac{x+z}{4}$$
  • To find the ratio $r$ of the geometric progression, factor out $x$: $$y = xr= x\frac{1+\frac {z}{x}}{4} \stackrel{\color{blue}{q = \frac{z}{x}}}{=} x\frac{1+q}{4}$$
  • Set up an equation for $\color{blue}{q = \frac{z}{x}}$ using $z = r^2 x$: $$z= x \left( \frac{1+q}{4}\right)^2 \Leftrightarrow q = \left( \frac{1+q}{4}\right)^2$$

Now, solve for $q$: $$16q = (1+q)^2 \Leftrightarrow q^2 - 14q + 1 = (q-7)^2 -48 = 0 \stackrel{q < 1}{\Rightarrow} \color{blue}{\boxed{q = 7 - 4\sqrt{3}}}$$

As already mentioned above by lulu, there are no integers $x,y,z$ satisfying the above given conditions.

trancelocation
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If one of $\,x,y,z\,$ were zero, then the other two would be zero as well by the GP condition, so $\,z/x\,$ would be undefined. Otherwise, it means that $\,xyz \ne 0\,$, then let $\,z/x = \lambda \lt 1 \iff z = \lambda x\,$, and the two conditions can be written as:

$$ \begin{align} 4 y = x + z = (1+\lambda)x \tag{1} \\ y^2 = xz = \lambda x^2 \tag{2} \end{align} $$

Squaring $(1)$, dividing by $(2)$, and canceling out the $\,y^2,x^2\,$ terms then gives $\require{cancel} \,16 \lambda = (1+\lambda)^2$ $\iff \lambda^2-14 \lambda+1=0\,$. The quadratic has two real irrational roots, one of which is $\,\lt 1\,$ and answers the question if the restriction to integers is dropped.

dxiv
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