# Is there an upper limit to Solve?

Posted 8 years ago
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 Hi everyoneThere 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 
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Posted 8 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 8 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.