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Does there exist $a\geq0$ such that $\lfloor a\rfloor,\lfloor a^2\rfloor,\lfloor a^3\rfloor,\ldots$ alternate between even and odd?

Problem adapted from the book Problem Solving Tactics

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    $\begingroup$ Bonus: does there exist a such that every floor(a^n) is congruent to n (mod p) for every p? Note: the actual question is for p=2. $\endgroup$ – William Nathanael Dec 7 '17 at 9:38
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    $\begingroup$ Is this really a puzzling question? I feel like it should be posted on math.se $\endgroup$ – Tweakimp Dec 7 '17 at 9:40
  • $\begingroup$ @WilliamNathanael Well, from the way I read this question, it should actually be "congruent to n+k (mod p)", since it's not defined whether floor(a) should be even or odd (At least, that's how I read it... perhaps wen did mean to start with even). $\endgroup$ – Lolgast Dec 7 '17 at 9:42
  • $\begingroup$ Tweakimp, nah, it ain't ordinary math question. @Lolgast Yup, missed that part. Still, I have no idea to approach this. Yet. $\endgroup$ – William Nathanael Dec 7 '17 at 9:46
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    $\begingroup$ @Tweakimp approved here $\endgroup$ – Wen1now Dec 7 '17 at 9:59
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The answer is

yes.

What follows is a sort of stream-of-consciousness solution; those skilled in the art may immediately recognize where I took a wrong turn. If you just want to see the answer, start reading at the end.

We can get numbers of the form $\left\lfloor a^n\right\rfloor$ as

solutions of a linear recurrence relation. Suppose e.g. the equation $t^2-2pt-q=0$ has distinct roots $a,b$ with $a>1$ and $0<b<1$; then the solutions to the recurrence $u_{n+2}=2pu_{n+1}+qu_n$ have the form $u_n=Aa^n+Bb^n$ for some $A,B$. If $B<0$ and $B$ is not too large then this will equal $\left\lfloor Aa^n\right\rfloor$.

Can we make this work? The conditions we need are:

$0<b<1<a$ as above. $A=1$ so that that last expression is the one we need. $B$ negative and not too large. $p,q$ integers with $q$ odd to make the recurrence relation yield integer results where alternate signs are equal.

Ah, no, this can't work because

if $A=1$ then I think the only way to make the surds go away is to have $B=1$ as well, and we need $B$ to be negative.

But perhaps it can be done

with a higher-order recurrence? We need one "big" root and the others all "small". If the thing corresponding to $A$ above is 1 (as it must be) then I think we again need the other coefficients to be 1, and we need the powers of the other roots to conspire to always add up to something negative.

That seems like it might be difficult or impossible. Here's another way. If

we instead make $b$ small and negative then the terms of the recurrence relation will alternate between being just below and just above $Aa^n$. So if we arrange for those always to be even and then take $A=B=1$ then everything may work. So perhaps try something like $t^2-4t-2$? (Even coefficients so that the corresponding recurrence relation perforce yields even values.) This has roots $2\pm\sqrt6$, and $-1<2-\sqrt6<0$ as we wanted, so $u_n=(2+\sqrt6)^n+(2-\sqrt6)^n$ is always an even integer and alternates between being just above and just below $(2+\sqrt6)^n$, whose floor therefore alternates between odd and even.

That last paragraph may be a bit opaque if one hasn't read the foregoing stuff, so here is a more self-contained version of the solution:

Suppose we let $a$ be one root of a quadratic equation with integer coefficients, and $b$ be the other root. Then the numbers $u_n=a^n+b^n$ satisfy a second-order recurrence relation with coefficients corresponding to those of the polynomial whose roots are $a,b$, and in particular they are all integers provided the coefficient of $t^2$ in that polynomial is 1. If the other two coefficients are even integers then our recurrence will look like $u_{n+2}=[\textrm{even}]u_{n+1}+[\textrm{even}]u_{n}$ forcing all the numbers it produces to be even. If $b$ is small then $a_n$ will be very close to the even integer $u_n$ (except maybe for small $n$); in this case, if $b>0$ then $a_n$ will always be just below an even integer, so its floor will always be odd, which is no use for us; but if $b<0$ then $a_n$ will alternate between being just below $u_n$ and being just above, so its floor will alternate between odd and even. OK, so let the quadratic equation be $t^2-2pt+2q=0$; its roots are $p\pm\sqrt{p^2-4q}$ so we had better have $q$ negative (so that the smaller root is negative) but small (to keep it close to zero). Let's try $t^2-4t-2=0$. The roots of this turn out to be $2\pm\sqrt6$; we get $-1<b<0$ and $a>1$ as we wanted, and indeed we can verify that $\left\lfloor(2+\sqrt6)^n\right\rfloor$ alternates in sign as required.

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