The discussion centers on finding all triples (a, b, c) of positive integers such that \(a^3 + b^3 + c^3 = (abc)^2\). The only confirmed solution is the triplet (1, 2, 3). Participants debated the necessity of distinct integers and explored the growth rates of the left and right sides of the equation, concluding that solutions are limited to small integers. Various algebraic approaches and a Python program were employed to search for additional solutions, but none were found beyond the confirmed triplet.
PREREQUISITESMathematicians, computer scientists, and enthusiasts in number theory seeking to understand the complexities of cubic equations and integer solutions.
Do we really need to prove a case such as ##(1,3,7)## analytically when direct substitution shows it does not satisfy the original equation?PAllen said:Now, by the argument @anemone posted, we have ##c^2\le 2b^3## and from above ##a^2b^2\lt 4/3c## for c>3. Putting these together, we end up with ##a^4b\lt 32/9##. This forces a=1, ##b\le 3##, thus b is 2 or 3. For b=2, this means c is 3 or 4 (by first equation in this paragraph). But 4 fails the derived relation ##a^2b^2>c##. So only 1,2,3 is possible for this case. For a=1, b=3, we have that c must be 4,5,6,or 7, by the same equation. But then from ##a^2b^2\lt4c/3##, we have that c can only be 7. Then by @anuttarasammyak mod 3 argument, this is excluded.
QED.
Of course not. It’s just avoiding direct test “as a matter of principle”. Obviously, reducing thousands of cases to test to 2 cases is more than adequate in practice.bob012345 said:Do we really need to prove a case such as ##(1,3,7)## analytically when direct substitution shows it does not satisfy the original equation?
Very interesting but could you please explain the divisibility requirement (##a^3+b^3## is divisible by ##c^2##) , I never understood that point. Thanks.PAllen said:Can't resist one more comment on this. @bob012345 asked the question "what changes past (1,2,3) to prevent another solution?". I can state now that the answer is for any bigger (a,b) value than (1,2) the range of c values that could possibly meet the divisibility requirement (##a^3+b^3## is divisible by ##c^2##) no longer has any overlap with where a real root must lie (c just below ##a^2 b^2##). At (1,2,3) these conditions overlap at c=3. Already at (a,b) = (1,3), the largest c possibly meeting the divisbility requirement is 5 (it doesn't, but it within range of being possible), while the real root is for c between 8 and 9. The failure of these regions to overlap grows ever larger as (a,b) get larger.
The RHS is an integer multiple of ##c^2##, so must the left. But ##c^3## is already multiple of ##c^2##, therefore ##a^3+b^3## must also be such a multiple. This, at minimum requires ##c^2 \le a^3+b^3##.bob012345 said:Very interesting but could you please explain the divisibility requirement (##a^3+b^3## is divisible by ##c^2##) , I never understood that point. Thanks.
Here the divisibility argument doesn't work because all it requires is ##c\le(a+b)## which is required for a solution anyway. Instead, simply solving for c leads to an expression which you can use to rule out c being an integer 'almost' the whole range of a and b (in fact you can show c < 2 almost all the (a,b) range). You are then left with just a few cases to test.bob012345 said:I presume using the same logic you could prove why there is only one solution for three positive integers to the relation? $$a+b+c=abc$$