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Is there an upper limit to Solve?

Paul Cleary
Posted 10 years ago

Hi everyone

There seems to be something unusual happening in this solve problem, when the power reaches 29 and above it either cant find a solution or there isn't a solution. I find it odd that this happens at this point, in some Solve problems I have done in the past has had many more digits than what is shown here. Can anyone shed any light on this? Thank you. 

Clear[a, b, c, p, m, k]; Do[m = 2 p + 1; 
 k = Flatten[{a, b, c} /. 
    Solve[a + b == m^p && b + c == (m + 1)^p && a + c == (m + 2)^p && 
      a > 0 && b > 0 && c > 0, {a, b, c}, Integers]]; 
 Print[k[[1]], " + ", k[[2]] " = ", m, "^", p]; 
 Print[k[[2]], " + ", k[[3]] " = ", m + 1, "^", p]; 
 Print[k[[1]], " + ", k[[3]] " = ", m + 2, "^", p]; 
 Print[], {p, 26, 29}]

675857535655680304849919087693057930899887829 + 2053352975677225647248849428897869261528900  = 53^26

2053352975677225647248849428897869261528900 + 1100089161598382798438690835729278647469252796  = 54^26

675857535655680304849919087693057930899887829 + 1100089161598382798438690835729278647469252796  = 55^26



97509545048771846906877632146531997884899466216 + 167523300201623773995913641696513925403268159  = 55^27

167523300201623773995913641696513925403268159 + 158715219758096757025533508534303691353204566977  = 56^27

97509545048771846906877632146531997884899466216 + 158715219758096757025533508534303691353204566977  = 57^27



14597663508899169035536438002129116461020344481753 + 7148085092341388610997016678517825551585410248  = 57^28

7148085092341388610997016678517825551585410248 + 23760369273139229384436648397631352217756817261368  = 58^28

14597663508899169035536438002129116461020344481753 + 23760369273139229384436648397631352217756817261368  = 59^28



a +  =  b59^29

b +  =  c60^29

a +  =  c61^29
POSTED BY: Paul Cleary
2 Replies
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Paul Cleary
Paul Cleary
Posted 10 years ago

Thank you Bill, that has sorted that little mystery, and indeed just removing the b condition gives solutions again. I will ponder this some more.

This ultimately boils down to

(2 k + 1)^k + (2 k + 2)^k > (2 k + 3)^k while k<=28
POSTED BY: Paul Cleary
Bill Simpson
Bill Simpson
Posted 10 years ago

Suppose

In[1]:= Reduce[{a + b == r, b + c == s, a + c == t}, {a, b, c}]

Out[1]= a == r/2 - s/2 + t/2 && b == r/2 + s/2 - t/2 && c == -(r/2) + s/2 + t/2

That result is simple enough that you can manually check that it will always be correct.

Now substitute in your constants.

In[2]:= p = 29;
m = 2 p + 1;
a == r/2 - s/2 + t/2 && b == r/2 + s/2 - t/2 && c == -(r/2) + s/2 + t/2 /. {r->m^p, s->(m+1)^p, t->(m+2)^p}

Out[4]= a == 2264609839938656962253544726423958389329925477265740 && 
 b == -1485905798391455475132628838090737282072934421601 && 
 c == 3686051234477280753804732628838090737282072934421601

It appears beginning at p=29 and at least up to 100 that b will be negative and this violates your condition that all solutions be positive.

Thus it appears there is no problem with Solve demonstrated thus far by this example.

The reason b goes negative is that it is the only item with a -t/2 and t > s > r. You might ponder why it waits until 29 for this to happen.
POSTED BY: Bill Simpson
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