## Maths challenge 6

Posted by Chris on June 30, 2011 – 8:04 pm

Following on from Karl’s smallest square twin problem, find a (I won’t insist on the smallest) cubic twin. i.e I want something that looks like 123123 = 50³ (that example doesn’t work).

As it has rather a lot of digits, I won’t insist on an explicit number.

July 1st, 2011 at 6:25 am

Well, that’s ruined my follow up question…

July 1st, 2011 at 12:19 pm

Hi Karl. Oh dear, I keep om doing that. Post your version and I’ll delete mine if yours is at least as good. Mine was written without too much thought – I revised it a few times because I kept on making it too hard.

July 1st, 2011 at 2:30 pm

No Chris, my question was simply, as you put it, what’s the lowest cubic twin… It appears to be considerably harder and larger than I imagined such a simple question to be! I take it you have approached this by a modulo route?

July 1st, 2011 at 3:31 pm

Hi Karl. Your questiion is much harder than mine (i.e.I don’t know of a slick way to get the smallest cubic twin). I don’t know a slick non Cadbury Number Crunchie way to get the smallest result. Even that route will be quite taxing. Factoring large numbers takes a looooong time.

July 2nd, 2011 at 12:04 am

Don’t worry, I haven’t got an answer yet, and, my last code/programme just failed, so back to the drawing board!

July 4th, 2011 at 6:43 pm

Looks like no-one is game for this. As with Karl’s Smallest Square problem, letting N be the half-twin, we want N(1 + 10

^{n}) = m^{3}. As before, we need 1+10^{n}to have a factor which is (at least) a cube. Let that factor be p^{3}. Then 1+10^{n}= kp^{3}. Then we need 1+10^{n}= 0 (mod p^{3}). But we know that10

^{φ}= 1 (mod p^{3}) where φ represents φ(p^{3}) = p^{2}(p-1). As before, p cannot be 2, 3 or 5. But φ(7^{3}) = 294 = 2*147, and sure enough 10^{147}= -1 (mod 7^{3}). So n = 147. If try p = 11 and 13, you simply get bigger numbers, but I haven’t tried further than that.So k = (1+10

^{147})/7^{3}= 29154518950437317784256559766763848396501457725947521865889212827988338192419825072886297376

09329446064139941690962099125364431486880466472303207.

I’m too lazy to finish this off. I’m satisfied to see such a ridiculously big number.

July 5th, 2011 at 3:35 am

Lazy mode subsided. That last number ≈ 2.9*10

^{144}. To pad it to 147 digits, we must multiply it by a cube, q^{3}, in the range 7^{3}/10 ≤ q^{3}< 7^{3}. So q = 4,5,6 will do. q = 4 => N =18658892128279883381924198250728862973760932944606413994169

09620991253644314868804664723032069970845481049562682215743

44023323615160349854227405248 and that’s the half twin

July 5th, 2011 at 8:20 am

Hi Chris,

I finally read the problem and the discussion, and I can tell you that you are wrong! The “cubic” twin problem is completely different than the “quadratic” one and my intuition tells me that in this setting we don’t have any solutions, at all. Where is the difference? In the quadratic problem it was enough to eliminate one of the prime factors of 10^n+1, while in the cubic case this is not the case – one needs to check if “sufficiently many” of the other prime factors appear in an even degree in the decomposition of 10^n+1. And this is extremely difficult to analyze! In simpler words, we need decomposition 10^n+1=A^3*B^2*C, such that B*C^2<10^n and good luck achieving it (Again, this is not enough, because the "massaging" factors like 2^3, or 3^3 may cause further problems).

July 5th, 2011 at 11:59 am

Hi slavy. How embarassing, you are absolutely right, unless there is an exeptionally special case where φ(p

^{3}) is usefully divisible by a large factor, the problem can’t be solved. I should have done the full analysis as in Karl’s Smallest Square problem; I would have ended up showing that N had too many digits, and so couldn’t do the job.NB Although I’ve focused on p prime, I know that it could be composite.

July 5th, 2011 at 12:38 pm

Hows about a cubic triplet!!!

We’d want NNN = m

^{3}=> N(1 + 10^{n}+ 10^{2n}) = m^{3}. At least we’re less likely to get the digits overflow problem. But we now need to find an n and f such that 1 + 10^{n}+ 10^{2n}= 0 (mod f). That looks impossible (or at least very unpromising) to me. We’d need e.g. 10^{n}= -x (mod f) and 10^{2n}= x-1 = x^{2}(mod f). Bearing in mind that we can offset -x and x-1 with multiples of f, we could spend a long time trying to find an (integer) solution for x. Offhand, this looks harder than the Pell’s equation problem. It’s not even immediately (if at all) apparent what to choose for f.I’ll pass on that problem.

July 5th, 2011 at 1:44 pm

Now I see why I have failed to write a programme, or why it fails…. the answer may not exist. I certainly couldn’t force a fit to your answer in post 6, so on that, I’m bugging out of this question too!