POTW Find Triplets of Positive Integers with Sum of Cubes

[bob012345]

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.

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]

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?

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.

 

[PAllen]

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 is 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.

 

[bob012345]

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.

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]

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.

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]

I presume using the same logic you could prove why there is only one solution for three positive integers to the relation? $$a+b+c=abc$$

 

[PAllen]

I presume using the same logic you could prove why there is only one solution for three positive integers to the relation? $$a+b+c=abc$$

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.