POTW Find Triplets of Positive Integers with Sum of Cubes

[bob012345]

It would seem the argument would have to be very special. Ending in 3 is certainly no good: $4(2)^4(3)^4~-~143$ is a perfect square. Ah, the key is simply the difference between consecutive perfect squares. The first expression under the radical is itself a perfect square. The difference between a perfect square and its predecessor is greater than 3 for all perfect squares greater than 4.

The polynomial with $n$ and $\alpha =b^2c^2$ has the property that when $n$ is integer $\alpha$ isn't and vice-versa which is a contradiction to the original assumptions of $n,b,c$. Graphically, the function for the positive root if we let x=b^2c^2 and have all real values,

$$n(x) = \frac{2x-3 + \sqrt{4x^2 -3}}{6}$$

asymptotically approaches the line from below $$ y= \frac{2x}{3} -\frac{1}{2}$$
which itself for any integer x is not an integer.

 

[anuttarasammyak]

Let me share my easy observation on modulus 3. With factorization of
a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
, the condition is written as
(a+b+c)((a+b+c)^2-3ab-3bc-3ca)=abc(abc-3)

In RHS
abc \equiv abc-3 (mod\ 3)
,
abc(abc-3) \equiv 0\ or \ 1(mod\ 3)

If it is 1, in LHS
a+b+c \equiv 1(mod\ 3) So
\{a,b,c\}\equiv \{-1,1,1\}(mod\ 3)

If it is 0
a+b+c \equiv 0 (mod\ 3) So
\{a,b,c\}\equiv\{-1,0,1\}(mod\ 3)
We may exclude the case of
a\equiv b\equiv c\equiv 0(mod\ 3)
for the concern mentioned in my previous post #32.

 

[PAllen]

Assume $a<b<c$, and let $a=n+1$ and $n≥0$.

$$(n+1)^3 + b^3 +c^3 = (n+1)^2b^2c^2$$ expanding
$$1 + 3n^2 +3n +n^3 + b^3 + c^3 = n^2b^2c^2 + 2nb^2c^2 + b^2c^2$$regrouping
$$n^3 + b^3 + c^3 -(nbc)^2 = -3n^2 + (2b^2c^2 - 3)n + b^2c^2 -1$$

It occurs to me that this whole approach is not correct. If the original equation is true for some (a,b,c), then the new (more complex) equation form(n,b,c) will be true. But a solution for making the RHS of the last equation above equal zero does not correspond to a solution of the original problem. Another way of saying this is that you have simply made a change of variables. For the new variables (n,b,c) a different equation is what corresponds to solutions in terms of the original variables. Thus I reject the validity of the whole approach.

 

[bob012345]

It occurs to me that this whole approach is not correct. If the original equation is true for some (a,b,c), then the new (more complex) equation form(n,b,c) will be true. But a solution for making the RHS of the last equation above equal zero does not correspond to a solution of the original problem. Another way of saying this is that you have simply made a change of variables. For the new variables (n,b,c) a different equation is what corresponds to solutions in terms of the original variables. Thus I reject the validity of the whole approach.

Thanks for pointing that out! I'm going to think more about it before I respond.

 

[bob012345]

One thing I thought was interesting about this problem is that for any integer $c$ there is always at least two integer solutions and two positive integer roots $(-c,0,c)$ and $(-c,c,c^4)$ but for any $c>3$ only those two positive roots . Any number for $c$ gives those roots as well. It's like there is some kind of mathematical phase change for $c>3$.

Also, a few more musings. Since the equation is symmetric in integers $(a,b,c)$ I wonder if it can be shown that the integers must be symmetric meaning $a$ and $c$ are equally spaced from $b$? Further, if that spacing must be one that leads immediately to $b=2$. I also wonder if the solution $(1,2,3)$ has anything to do with the fact that these numbers have the property that $a+b+c=abc$?

 

[anuttarasammyak]

I also wonder if the solution (1,2,3) has anything to do with the fact that these numbers have the property that a+b+c=abc?

Ref.#52 the formula of condition is written as
(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=abc(abc-3)
which might help checking your conjecture.

 

[bob012345]

Ref.#52 the formula of condition is written as
(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=abc(abc-3)
which might help checking your conjecture.

Thanks. How?

BTW, I noticed the original equation can be expressed as

$$a^2(a - b^2c^2) = -(b^3 + c^3)$$ and also

$$b^2(b - a^2c^2) = -(a^3 + c^3)$$ and

$$c^2(c - a^2b^2) = -(a^3 + b^3)$$

suggesting reciprocal relations $a < b^2c^2, b < a^2c^2, c < a^2b^2$

 

[anuttarasammyak]

I just see a+b+c and abc appear in the formula but have no idea how to do it by myself, I am afraid to say.

Your observation is confirmed by the relation
\frac{a}{b^2c^2}+\frac{b}{c^2a^2}+\frac{c}{a^2b^2}=1

 

[bob012345]

I just see a+b+c and abc appear in the formula but have no idea how to do it by myself, I am afraid to say.

Your observation is confirmed by the relation
\frac{a}{b^2c^2}+\frac{b}{c^2a^2}+\frac{c}{a^2b^2}=1

If one could start with the relation $a+b+c=abc$ which only works over the positive integers for $(1,2,3)$ and manipulate it to get the equation, I think that would prove it?

 

[PAllen]

Since there remains interest on how to improve the initial correct solution to reduce cases to test, I decided to post a long winded deduction based on the analysis behind my efficient search program combined with the key insight of the posted solution that I failed to come up with. With this combination, I can arrive at the conclusion that there are exactly two cases to test: (1,2,3) and (1,3,7). All others are ruled out by analysis. If I follow correctly, the second of these is then excluded by @anuttarasammyak mod 3 analysis in post #52. This leaves just one example to test, which works. After this, the remaining challenge is to come up with a less verbose argument to this "best result".

I should also note that as stated, the solution posted by @anemone leaves quite a large number of cases to test. Switching to my preferred notation of $a\le b\le c$, you have a=1, $b\le 18$, and $c\le 108$. That is really untestable in practice without a computer program.

The first step is justify what @bob012345 has stated: assuming first that you have positive integers $1 \le a \le b \le c$ it can be concluded that you must actually have $1 \le a \lt b \lt c$. There are several ways to show this. I give what I think are arguments requiring minimum background knowledge.

First, consider the possibility $a=b=c \gt 0$. Then we have the claim $3c^3=c^6 \Rightarrow 3=c^3$, impossible for integer c.

Next consider the possibility $a \lt b=c$. Then we have $a^3+2c^3=a^2c^4$, and then $a^3=c^3(a^2c-2)$. But then we must have $a^2c-2$ being in the open interval (0,1), which is impossible.

Finally, consider the possibility $a=b\lt c$. We have $2a^3+c^3=a^4c^2$, then $c^3/a^3=ac^2-2$. Since the RHS is an integer, so is the LHS, and an integer can only have an integer or irrational cube root. The latter is excluded (else c/a would be irrational), thus we must have c=ka for some positive integer k. With substitution and algebra we get $a^3=k+2/k^2$. But this can only be integer if k=1, which contradicts the assumption that $c\gt a$. Thus this case is also impossible.

Thus we know $1 \le a \lt b \lt c$ and $c\ge 3$.

Now we show that for a solution it must be true that $c \lt a^2b^2 \lt (4c/3)$ (the second inequality only being true for c>3). The first of these inequalities is easy. The main length of this argument is the last inequality, which is crucial for reducing the number of cases to test.

So, assume $a^2b^2\le c$. Then $a^3+b^3+c^3=(a^2 b^2)c^2 \le c^3$. This is impossible, so the first inequality is proven.

So now we assume $a^2b^2\ge(4c/3)$ along with all other assumptions. This gives $(3/4) \ge c/(a^2b^2)$. Negating this and adding 1 gives: $1/4 \le 1- c/a^2b^2$. From the target equation we can write $a/b^2+b/a^2=(1-c/a^2b^2)c^2$. Then, $a/b^2+b/a^2\ge c^2/4$. For any choice of b, the LHS can be maximized by letting a=1 (I am not sure how to prove this without calculus, but it is pretty easy with calculus - there is one extremal, it is a minimum; then the greatest value is at the extreme of the range). Thus we must have (given starting assumptions): $1/b^2+b \ge c^2/4$. Given b < c, it is easy to see this must be false for c>3. We have thus established what was desired.

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 $a^4b\le 32/9$. 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.

 

[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.