Crikey Math

Having fun exploring mathematics

  • Home
  • About
  • Blog
  • Contribute
en English
af Afrikaanssq Albanianam Amharicar Arabichy Armenianaz Azerbaijanieu Basquebe Belarusianbn Bengalibs Bosnianbg Bulgarianca Catalanceb Cebuanony Chichewazh-CN Chinese (Simplified)zh-TW Chinese (Traditional)co Corsicanhr Croatiancs Czechda Danishnl Dutchen Englisheo Esperantoet Estoniantl Filipinofi Finnishfr Frenchfy Frisiangl Galicianka Georgiande Germanel Greekgu Gujaratiht Haitian Creoleha Hausahaw Hawaiianiw Hebrewhi Hindihmn Hmonghu Hungarianis Icelandicig Igboid Indonesianga Irishit Italianja Japanesejw Javanesekn Kannadakk Kazakhkm Khmerko Koreanku Kurdish (Kurmanji)ky Kyrgyzlo Laola Latinlv Latvianlt Lithuanianlb Luxembourgishmk Macedonianmg Malagasyms Malayml Malayalammt Maltesemi Maorimr Marathimn Mongolianmy Myanmar (Burmese)ne Nepalino Norwegianps Pashtofa Persianpl Polishpt Portuguesepa Punjabiro Romanianru Russiansm Samoangd Scottish Gaelicsr Serbianst Sesothosn Shonasd Sindhisi Sinhalask Slovaksl Slovenianso Somalies Spanishsu Sudanesesw Swahilisv Swedishtg Tajikta Tamilte Teluguth Thaitr Turkishuk Ukrainianur Urduuz Uzbekvi Vietnamesecy Welshxh Xhosayi Yiddishyo Yorubazu Zulu

(2^(p-1)-1)/p is not a square for p >7 and prime

September 9, 2017 By Gary Ernest Davis

@jamestanton  raised the following (implicit) question on Twitter on September 8, 2017:

In the spirit of playing around and not thinking too hard, just computing to see what might be going on, I wrote a short piece of Mathematica® code that will run through the primes p from the 2^{nd} through the 1000^{th} and check when  (2^{p-1}-1)/p is a square:

n = 2;
L = {};
While[n <= 1000,
If[IntegerQ[Sqrt[(2^(Prime[n] – 1) – 1)/Prime[n]]],
L = {L, Prime[n]}]; n++]
L = Flatten[L]

with the result {3, 7}.

So no more primes p >7 for which (2^{p-1}-1)/p is a square.

To continue in this vein I just let Mathematica® keep on running through the primes greater than 7, stopping only if it finds a prime p >7 for which (2^{p-1}-1)/p is a square:

n = 5;
While[IntegerQ[Sqrt[(2^(Prime[n] – 1) – 1)/Prime[n]]] == False, n++]
n
Prime[n]

I kept this running, with no positive result and aborted at which point I found n = 43984, and the n^{th} prime to be 532159.

So by now I was thinking that maybe p = 3 and p = 7 is it for primes p for which (2^{p-1}-1)/p is a square.

If that is indeed the case, then it would be sort of nice to figure where in the non-square numbers (2^{p-1}-1)/p appears.

Now, the n^{th} non-square number is \lfloor \sqrt{n} + n + 1/2 \rfloor.

As a reminder: \lfloor x \rfloor is the greatest integer \leq x .

You can find this fact, for example, in Jim Tanton’s book Mathematics Galore!  in Chapter 4.

So the idea is to find an integer n for which (2^{p-1}-1)/p =\lfloor \sqrt{n} + n + 1/2 \rfloor when p > 7 is prime.

If we drop the floor function temporarily we get a quadratic equation in n and the smaller of the roots – which a bit of reflection tells us is the correct root – is (2^{p-1}-1)/p -\sqrt{(2^{p-1}-1)/p -1/4} .

Of course this root is NOT an integer, and again a bit of reflection tells us that we want the ceiling of this root, so we set n:= \lceil(2^{p-1}-1)/p -\sqrt{(2^{p-1}-1)/p -1/4} \rceil .


So now the idea is to prove that if p >7 is prime and

n:= \lceil(2^{p-1}-1)/p -\sqrt{(2^{p-1}-1)/p -1/4} \rceil

then

(2^{p-1}-1)/p =\lfloor \sqrt{n} + n + 1/2 \rfloor

This would tell us that (2^{p-1}-1)/p is non-square forp >7 , and gives us something concrete with which to work.

http://www.crikeymath.com/wp-content/uploads/2017/09/crikey.wav

 

 

Filed Under: Uncategorized

Recent Posts

  • Does the DNA test say guilty or not guilty?
  • The Turtleback Diagram for Conditional Probability
  • Numbers that, base 2, are blindingly obviously divisible by 3
  • A mysterious sequence of 1s and 2s
  • Does one have to be a genius to do mathematics? Neither necessary nor sufficient.
  • Alexander Bogomolny
  • Binary disjoints of an integer
  • A cycle of length 5 for iterated squaring modulo a prime
  • The diagonal of the Wythoff array
  • Leonardo Bigollo, accountant, son of Bill Bonacci

Archives

Copyright © 2023 · Dynamik Website Builder on Genesis Framework · WordPress · Log in