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 Heiro's Wager (Posted on 2010-09-08)

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)

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 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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