The discussion revolves around finding all triples (a, b, c) of positive integers such that the equation \(a^3 + b^3 + c^3 = (abc)^2\) holds. Participants explore various approaches, assumptions, and potential solutions related to this mathematical problem.
Participants do not reach a consensus on whether there are solutions beyond the known triplet (1, 2, 3), and multiple competing views remain regarding the conditions and implications of the equation.
There are unresolved assumptions about the distinctness of integers a, b, and c, and the limitations of the search range for potential solutions. The discussion also highlights the complexity of proving the existence or non-existence of solutions.
Readers interested in number theory, mathematical problem-solving, or those exploring the properties of cubic equations may find this discussion relevant.
for a in range(100):
for b in range(a,100):
for c in range(b,100):
abc = a * b * c
abc2 = abc * abc
a3 = a * a * a
b3 = b * b * b
c3 = c * c * c
sum = a3 + b3 + c3
if sum == abc2:
print("a=%d / b=%d / c=%d\n"%(a,b,c))ambiguity alert: Are a,b, and c required to be different positive integers? I assume not.anemone said:Find all triples (a, b, c) of positive integers such that ##a^3+b^3+c^3=(abc)^2##.
PAllen said:Where's your proof that there are no solutions outside of your search range?
It reads "find ALL"jedishrfu said:@anemone asked for three numbers
Let c be variable xPAllen said:Where's your proof that there are no solutions outside of your search range?
That wasn't stated but it is easy to show whether that must be the case or not. Assume they are the same and see if the real root is integer or not.PAllen said:ambiguity alert: Are a,b, and c required to be different positive integers? I assume not.
Your partial solution and poem are definitely amusing but the above argument is wholly inadequate. Even if you ban duplicates, the following is true: ##for~every~c \gt 3, there~exist~triples~a\lt b\lt c## such that sum of cubes is greater than the product of squares, and other triples where it is less. Thus no matter how large c is, there is crossover region. Once the problem is properly solved, I can give a simple formula for generating near equality triples for arbitrary c. In problems like this, where all solutions are requested, the proper demonstration of that is typically where 'all the real math' is.jedishrfu said:Pr
No proof here. @anemone asked for three numbers which I took to be three distinct numbers.
However, by inspection one can see that if the numbers were 0, 1, n where n can vary from 1 to infinity the left side value will increase much faster than the right side value implying that the answer must be in small numbers.
Also to be solvable it must be limited to what we can do manually on paper or mentally in our heads.
Being a robot, I initially searched on a much larger range but then realized that 1-100 would be sufficient and not give away the solution. Being a mentor, I tried to disguise my solution like Penn and Teller on Fool Us in clever banter or maybe not so clever banter.
But there is no (integer) real root thus no solution in terms of integer triples in that case for any ##a##.Office_Shredder said:The solution set is not obviously bounded. If ##a=b## and ##c=1##, then the right hand side grows faster.
d=0
rat=10
for c in range(4,10000001):
for a in range(1,int(c**.25)+1):
b=int(c**.5/a)+1
while 1:
L=a**3 + b**3 + c**3
R=(a*b*c)**2
if R==L: d=1
nrat=R/L
if nrat<rat:
rat=nrat
sa=a
sb=b
sc=c
if R >= L : break
b=b+1
if d==1: break
if d==1: break
print(sa,sb,sc)
print(rat)I’m not sure even this can be required without further justification. You could have c=42, p=63, q=28. p is < 2c, q < ##c^2##, both are factors of ##c^2##, but neither are factors of c. And, of course, the product is ##c^2##. There may well be an argument to exclude this possibility, but it has to be givenanuttarasammyak said:The formula (2) in post #18 gives the cases of
Case 1
a+b\equiv 0(mod\ c^2)
Case 2
a^2-ab+b^2 \equiv 0(mod\ c^2)
Case 3
a+b\equiv 0(mod\ p),
a^2-ab+b^2 \equiv 0(mod\ q)
where ##pq=c^2## where p is a devisor of c.
You've lost me here. Where does ##-n^2b^2c^2## come from?bob012345 said:My attempt is here.
Given ##a<b<c## let ##a=n+1## where ##n≥0##
$$(n+1)^3 + b^3 +c^3 = (n+1)^2b^2c^2$$ expanding
$$1 + 3n^2 +3n +n^3 + b^3 + c^3 -n^2b^2c^2 = b^2c^2(n^2 + 2n + 1)$$
Thanks. p is a divisor of c^2, not of c, my bad.PAllen said:I’m not sure even this can be required without further justification. You could have c=42, p=63, q=28. p is < 2c, q < c2, both are factors of c2, but neither are factors of c. And, of course, the product is c2. There may well be an argument to exclude this possibility, but it has to be given
Sorry, I corrected a double counting of that term but it comes from the first term of $$b^2c^2(n^2 + 2n + 1)$$ from the RHS just moved to the LHS. It should not have been on the LHS at that point and has been corrected. Thanks.PAllen said:You've lost me here. Where does ##-n^2b^2c^2## come from?
This argument is not valid as stated. All you have shown is that n=0 is part of a possible solution. The fact that many choices of n,b,c don't work, and some make the LHS negative, proves nothing. You need an argument that no such choices work, if n>0.bob012345 said:$$n^3 + b^3 + c^3 -(nbc)^2 = nb^2c^2 + (n+1)(b^2c^2-3n) -1$$
Notice that the LHS is just our original equation with ##n## instead of ##a##. Also, notice that the RHS is always positive since ##b^2c^2>3n## yet the LHS can be negative or positive. Demanding ##n=0## removes the inconsistency thus the smallest integer is ##a=1##.
I don't see the inconsistency here. ##n = a-1## is fixed. It's not a free variable.bob012345 said:Notice that the LHS is just our original equation with ##n## instead of ##a##. Also, notice that the RHS is always positive since ##b^2c^2>3n## yet the LHS can be negative or positive. Demanding ##n=0## removes the inconsistency thus the smallest integer is ##a=1##.
At the risk of making myself look stupid (not hard), could someone kindly explain something from the proof in the Post #29 Spoiler.anemone said:However, here is one brilliant solution of someone that I want to share:
WLOG, let ##a\ge b \ge c##. Note that
##(abc)^2 \le 3a^3 \implies a\ge \dfrac{(bc)^2}{3}##
On the other hand, we have
##a^2 \mid b^3+c^3 \implies a^2\le2b^3##
Combining both bounds gives ##\dfrac{(bc)^4}{9}\le 2b^3 \implies bc^4\le 18##. Thus, ##c=1## and ##b\le 18##.
Check the remaining cases manually, the answers are the six permutations of ##(3, \, 2,\,1)##.