POTW Find Triplets of Positive Integers with Sum of Cubes

[PeroK]

At the risk of making myself look stupid (not hard), could someone kindly explain something from the proof in the Post #29 Spoiler.

It seems that we jump from $a\ge b \ge c$ to $(abc)^2 \le 3a^3$.

You combine that inequality with $(abc)^2 = a^3 + b^3 + c^3 \le 3a^3$.

 

[anuttarasammyak]

However, here is one brilliant solution of someone that I want to share:

Smart! Thanks for the sharing.

I would like to add some observation. We can introduce
a=Ag,b=Bg,c=Cg
where g is common divisor among the three and A,B,C are coprime to others. The original problem is expressed as
A^3+B^3+C^3=(ABC)^2g^3
By this answer g=1 which brings the original formula and now we know a, b and c are coprime. And
b<a<b^2c^2
Those may reduce candidates of a and b to be tested to some extent.

 

[PeroK]

I would like to add some observation. We can introduce
a=Ag,b=Bg,c=Cg
where g is common devisors among the three and A,B,C are coprime to others.

Not necessarily: $a = 2, b = 6, c = 15$ have no common factor, but they are not pairwise coprime.

The original problem is expressed as
A^3+B^3+C^3=(ABC)^2g^3
By this answer g=1

Why?

 

[PeroK]

PS it's true that if $a, b, c$ is a solution, then $ka, kb, kc$ is not a solution for $k > 1$.

 

[bob012345]

I don't see the inconsistency here. $n = a-1$ is fixed. It's not a free variable.

The only condition I assumed is that $a<b<c$ so it is not fixed.

 

[PeroK]

I set the condition $The only condition I assumed is that$a<b<c## so it is not fixed.

Your inequality only rules out cases where the signs of the LHS and RHS are different. It doesn't rules out cases where both the RHS and LHS are $> 0$.

 

[anuttarasammyak]

Not necessarily: a=2,b=6,c=15 have no common factor, but they are not pairwise coprime.

a, b, c is the number satisfying the formula in OP. (2,6,15) does not.
If there exists a common divisor of a and b, it must be also a divisor of c.

 

[PeroK]

Common divisor of a and b must be also a divisor of c.

Okay.

 

[anuttarasammyak]

I would say if (2,6,15) is a solution, 15 should be divisible by 2 which is common divisor of 2 and 6.
2^3\equiv 0(mod \ 2)
6^3\equiv 0(mod \ 2)
(2*6*15)^2\equiv 0(mod \ 2)
why not
15^3 \equiv 0(mod \ 2)
if it is a right solution.

Solution (a,b,c) should have no common factors, e.g. say a is even, b and c are odd. Say a is multiple of 3, b and c are not divisible by 3, $\{2*3^2, 5*7^4, 11*13^3\}$
.

 

[bob012345]

On the other hand, we have

$a^2 \mid b^3+c^3 \implies ...$

What does this step mean? $a^2 \mid b^3$?

 

[PAllen]

What does this step mean? $a^2 \mid b^3$?

It means $b^3+c^3$ is divisible by $a^2$. Note, this derivation used the reverse convention of everyone else on this thread: a is largest rather than c is largest.

 

[anuttarasammyak]

As for case 3, in my post #8 I made
p=q=c
Now I find that other cases should be investigated for possible c of non prime number.

Say integers $0<a\leq b \leq c$ sasisfy
a^3+b^3+c^3=(abc)^2
Let
a+b=pm
a^2-ab+b^2=(a+b)^2-3ab=qn
where $pq=c^2, m,n=1,2,...$

We have some conditions on largeness of m,n
m&lt;2c/p,\ n&lt;p

We get the formula
9(mn+c)=(p^2m^2-qn)^2

I am stuck here.

 

[bob012345]

Thanks everyone for the overwhelming participation. I want to encourage for those who have "partially" solved the question right but cannot prove those are the unique answers to have further attempt and I wish you success in finding the proof.

Thanks, I will since spent so much effort already.

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.

I'll try a different approach using the same basic idea. Hopefully it is a stronger argument. It still needs a bit of work.

Spoiler

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
$$n^3 + b^3 + c^3 -(nbc)^2 = -3n^2 + (2b^2c^2 - 3)n + b^2c^2 -1$$
The LHS is the original equation which we are assuming equals zero for the right numbers of n,b,c thus

$$ 3n^2 - (2b^2c^2 - 3)n - b^2c^2 + 1 = 0$$ This gives the only positive solution for $n$

$n = \frac{1}{6}\left(\sqrt{4b^4c^4 -3} + 2b^2c^2 -3\right)$

Just looking at the square root we can see it will always be irrational since it can never be a perfect square. [EDIT: thanks to @PAllen who pointed out it can]. Then this can be written as$$ n = \frac{1}{6}\left(2b^2c^2 \sqrt{1 -\frac{3}{4(b^2c^2)^2}} + 2b^2c^2 -3\right)$$ or
$$ n = \frac{b^2c^2}{3}\left( 1 + \sqrt{1 -\frac{3}{4(b^4c^4)}} \right) -\frac{1}{2}$$

This shows there can be no positive integer values for n for any integer $b,c$. Since we assumed an integer $n$ for the LHS and always have a non-integer $n$ for the RHS for any positive $b,c$ so we must conclude that $n$ must be zero and $a=1$. This shows that at least one of the three must be 1.

Now we have

$$1 + b^3 + c^3 =b^2c^2$$ apply the same logic to $b$ letting $b=n+2$ $n≥0$.

$$1 + (n+2)^3 + c^3 = (n+2)^2c^2$$ expanding

$$1 + n^3 +3(2)n^2 + 3(2^2)n + 2^3 + c^3 = (n^2 + 4n +4)c^2$$ regrouping$$1 + n^3 + c^3 - (nc)^2= 4c^2 + 4nc^2 -6n^2 -12n -8$$ setting the LHS to zero we get

$$6n^2 -4(c^2 - 3)n -4c^2 +8 = 0$$ giving the only positive solution

$n = \frac{1}{3}\left(\sqrt{c^4 -3} + c^2 -3\right)$
which is never an integer so there are no other integers for $b$ except $b=2$. This shows one of the values must be $2$ leaving only $c=3$.

[\SPOILER]

 

[PAllen]

I'll try a different approach using the same basic idea. Hopefully it is a stronger argument. 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

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

It seems you lost a factor of 2 here.

The LHS is the original equation which we are assuming equals zero for the right numbers of n,b,c thus

$$ 3n^2 - (b^2c^2 - 3)n - b^2c^2 + 1 = 0$$

continuing here

This gives the only positive solution for $n$

$n = \frac{1}{6}\left(\sqrt{4b^4c^4 -3} + 2b^2c^2 -3\right)$

It looks like you gained back the missing 2, but you are missing an important term under the radical.

"Just looking at the square root we can see it will always be irrational since it can never be a perfect square"

You need a clear argument here. I had a false proof involving alleged impossibility of an expression being a perfect square - but poked holes in before 'publishing'. Note, in particular, that a 4th power of an integer minus an odd prime can certainly be a perfect square, e.g. 16-7.

 

[bob012345]

It seems you lost a factor of 2 here.

continuing hereIt looks like you gained back the missing 2, but you are missing an important term under the radical.

What term? I don't think I am missing a term under the radical according to Wolfram Alpha?

"Just looking at the square root we can see it will always be irrational since it can never be a perfect square"

You need a clear argument here. I had a false proof involving alleged impossibility of an expression being a perfect square - but poked holes in before 'publishing'. Note, in particular, that a 4th power of an integer minus an odd prime can certainly be a perfect square, e.g. 16-7.

Thanks. The term under the radical always ends in either 1 or 7. The term $(bc)^4$ will always end in either 1,5 or 6. Multiply it by 4 and it will end in either 0 or 4. Subtract 3 and it will end in 1 or 7. I believe any number ending in 1 or 7 will have an irrational square root.

 

[PAllen]

What term? I don't think I am missing a term under the radical according to Wolfram Alpha?

The whole expression under the radical starts out $(2b^2c^2-3)^2-12$ This generates a term $-12b^2c^2$ which I don't see how can simplify away.

Thanks. The term under the radical always ends in either 1 or 7. The term $(bc)^4$ will always end in either 1,5 or 6. Multiply it by 4 and it will end in either 0 or 4. Subtract 3 and it will end in 1 or 7. I believe any number ending in 1 or 7 will have an irrational square root.

81

 

[bob012345]

The whole expression under the radical starts out $(2b^2c^2-3)^2-12$ This generates a term $-12b^2c^2$ which I don't see how can simplify away.

There was a typo in my post since I had rewritten it over and over but equation I solved was

$$3n^2 - (2b^2c^2 - 3)n - b^2c^2 + 1 = 0$$

so the solution is $$n = \frac{(2b^2c^2 - 3) ±\sqrt{(2b^2c^2 - 3)^2 - 4(3)(- b^2c^2 + 1)}}{6}$$

$$n = \frac{(2b^2c^2 - 3) ±\sqrt{4b^4c^4 -12b^2c^2 +9 +12b^2c^2 -12}}{6}$$
$$n = \frac{(2b^2c^2 - 3) ±\sqrt{4b^4c^4 -3)}}{6}$$

81

$\sqrt{4(27)^{4}(28)^{4}-3}= 1143072$

Ok, any suggestions?

 

[PAllen]

There was a typo in my post since I had rewritten it over and over but equation I solved was

$$3n^2 - (2b^2c^2 - 3)n - b^2c^2 + 1 = 0$$

so the solution is $$n = \frac{(2b^2c^2 - 3) ±\sqrt{(2b^2c^2 - 3)^2 - 4(3)(- b^2c^2 + 1)}}{6}$$

$$n = \frac{(2b^2c^2 - 3) ±\sqrt{4b^4c^4 -12b^2c^2 +9 +12b^2c^2 -12}}{6}$$
$$n = \frac{(2b^2c^2 - 3) ±\sqrt{4b^4c^4 -3)}}{6}$$

ok, great.

$\sqrt{4(27)^{4}(28)^{4}-3}= 1143072$

Ok, any suggestions?

No, not really. I gave up on an approach involving proving an expression with subtraction couldn't be a perfect square - because I couldn't come up with a valid argument. (I was playing with the rational roots theorem instead, but, at a key point I had a similar issue to yours here, and gave up).

 

[bob012345]

ok, great.

No, not really. I gave up on an approach involving proving an expression with subtraction couldn't be a perfect square - because I couldn't come up with a valid argument. (I was playing with the rational roots theorem instead, but, at a key point I had a similar issue to yours here, and gave up).

BTW, I was incorrect, $\sqrt{4(27)^{4}(28)^{4}-3}≠ 1143072$ it only appeared so when rounded. Also, Wolfram Alpha says there are no integer values but I'd like to prove it myself. Thanks for your help!

 

[PAllen]

BTW, I was incorrect, $\sqrt{4(27)^{4}(28)^{4}-3}≠ 1143072$ it only appeared so when rounded. Also, Wolfram Alpha says there are no integer values but I'd like to prove it myself. Thanks for your help!

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.

 

[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.