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Heiro's Wager (Posted on 2010-09-08) Difficulty: 3 of 5

Heiro: Good morning, Archimedes! I bet you 1000 drachma that you can’t answer a simple maths problem!

Archimedes: You’re on.

Heiro: Did you know that every sum of consecutive cubes is also a difference of squares?

Archimedes: Indeed, as everyone knows, each and every sum of consecutive cubes is also a difference of consecutive squares. Was that the question?

Heiro: Not so fast, Archimedes! What if the sum and difference are a prime number?

Archimedes (yawning): Unlikely, since any sum of cubes has at least two factors…

Heiro: Let’s make it a difference of consecutive cubes, then?

Archimedes: There is no reason why such a difference should not be prime, as often as occasion demands. For example, 2^3-1^3 = 7. Have I won yet?

Heiro: I'm just working up to it. What about a difference of consecutive cubes, which is not only prime, but is also a sum of consecutive squares?

Archimedes: Too easy! 61 = 5^3-4^3=5^2+6^2. Can I have my money now?

Heiro: Very well, then, here's my question – just to make it interesting, let’s also stipulate that a side of the larger cube must have at least 4 distinct prime factors – so it must be at least 2*3*5*7 = 210, say? Just how big would that cube have to be?

Archimedes: Easy again! – or is it? Wait a minute…

What is the answer to Heiro’s simple problem?

See The Solution Submitted by broll    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts not getting far | Comment 1 of 3

Programmed to find solutions with 3 or more factors in the larger cube:

   10   for Strt=1 to 1000000
   20    Diff=(Strt+1)^3-Strt^3
   30    if prmdiv(Diff)=Diff or prmdiv(Diff)=0 then
   40      :St2=int(sqrt(Diff/2))
   50      :if St2^2+(St2+1)^2=Diff then
   60         :if fnFactor(Strt+1)>2 then
   80           :print Strt;Strt+1,St2;St2+1,Diff
  100   next
  999   end
 1000   fnFactor(Num)
 1010     S$="":N=abs(Num):Ct=0
 1020     if N>0 then Limit=sqrt(N):else Limit=0
 1030     if Limit<>int(Limit) then Limit=int(Limit+1)
 1040     Dv=2:gosub *DivideIt
 1050     Dv=3:gosub *DivideIt
 1060     Dv=5:gosub *DivideIt
 1070     Dv=7
 1080     loop
 1090      if Dv>Limit then goto *Afterloop
 1100      gosub *DivideIt:Dv=Dv+4 '11
 1110      gosub *DivideIt:Dv=Dv+2 '13
 1120      gosub *DivideIt:Dv=Dv+4 '17
 1130      gosub *DivideIt:Dv=Dv+2 '19
 1140      gosub *DivideIt:Dv=Dv+4 '23
 1150      gosub *DivideIt:Dv=Dv+6 '29
 1160      gosub *DivideIt:Dv=Dv+2 '31
 1170      gosub *DivideIt:Dv=Dv+6 '37
 1180      if inkey=chr(27) then S$=chr(27):end
 1190    endloop
 1200    *Afterloop
 1210    if N>1 then S$=S$+str(N):Ct=Ct+1
 1220   'print S$
 1230   return(Ct)
 1240 
 1250   *DivideIt
 1255   Did=0
 1260    loop
 1270     Q=int(N/Dv)
 1280     if Q*Dv=N and N>0 then
 1290       :N=Q:S$=S$+str(Dv):Did=1
 1300       :if N>0 then Limit=sqrt(N):else Limit=0:endif
 1310       :if Limit<>int(Limit) then Limit=int(Limit+1):endif
 1320      :else
 1330      :goto *Afterdo
 1340     :endif
 1350    endloop
 1360    *Afterdo
 1365   Ct=Ct+Did
 1370    return

this program finds, looking at cubes up to a million cubed, only a situation where the larger cube has only three unique prime factors:

427285^3 - 427284^3 = 523314^2 + 523315^2 = 547716131821

where 427285 = 5 * 97 * 881. The cube of course has the same prime factors, just repeated three times each.

The numbers have to be higher if they exist at all.


  Posted by Charlie on 2010-09-08 18:39:41
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