MHB Solving the System of Equations for $a^2+b^2+c^2$

[anemone]

$a,\,b,\,c$ are integers that satisfy the system of equations below:

$a^2b+b^2c+c^2a=2186$

$ab^2+bc^2+ca^2=2188$

Evaluate $a^2+b^2+c^2$.

 

[Opalg]

$a,\,b,\,c$ are integers that satisfy the system of equations below:

$a^2b+b^2c+c^2a=2186$

$ab^2+bc^2+ca^2=2188$

Evaluate $a^2+b^2+c^2$.

Partial solution:
[sp]If $x$ is the smallest of the three numbers $a,\,b,\,c$ then $3x^3 \leqslant 2186$. And if $y$ is the largest of the three numbers then $3y^3 \geqslant 2188$. Having seen that, I noticed that $3\cdot9^3 = 2187.$ It follows that $x<9$ and $y>9.$

The next step was pure guesswork. I noticed that if $a = b-1$ and $c=b+1$ then $$a^2b+b^2c+c^2a = (b-1)^2b + b^2(b+1) + (b+1)(b^2-1) = 3b^3 - 1,$$ $$ab^2+bc^2+ca^2 = (b-1)b^2 + b(b+1)^2 + (b^2-1)(b-1) = 3b^3+1.$$

It follows that $(a,b,c) = (8,9,10)$ is a solution, with $a^2+b^2+c^2 = 64+81+100 = 245.$

Of course, that can't count as a proper solution, because it relies on a lucky guess. More seriously, it does not show that the solution is unique.[/sp]

 

[anemone]

Partial solution:
[sp]If $x$ is the smallest of the three numbers $a,\,b,\,c$ then $3x^3 \leqslant 2186$. And if $y$ is the largest of the three numbers then $3y^3 \geqslant 2188$. Having seen that, I noticed that $3\cdot9^3 = 2187.$ It follows that $x<9$ and $y>9.$

The next step was pure guesswork. I noticed that if $a = b-1$ and $c=b+1$ then $$a^2b+b^2c+c^2a = (b-1)^2b + b^2(b+1) + (b+1)(b^2-1) = 3b^3 - 1,$$ $$ab^2+bc^2+ca^2 = (b-1)b^2 + b(b+1)^2 + (b^2-1)(b-1) = 3b^3+1.$$

It follows that $(a,b,c) = (8,9,10)$ is a solution, with $a^2+b^2+c^2 = 64+81+100 = 245.$

Of course, that can't count as a proper solution, because it relies on a lucky guess. More seriously, it does not show that the solution is unique.[/sp]

Thank you Opalg for participating and also your solution, the educated guess is spot on!

Solution of other:

Spoiler

If we subtract the first equation from the second, we get:

$ab(b-a)+bc(c-b)+ac(a-c)=2$

Now setting $a=b$ turns the LHS into $a^2(a-a)+ac(c-a)+ac(a-c)$, which after simplifying equals to 0. This means $a-b$ is a factor of the LHS.

Similarly, $b-c$ and $c-a$ are also factors. Therefore we get the factored equation $(a-b)(b-c)(c-a)=2$.

Note that $(a-b)+(b-c)+(c-a)=0$, the ordered triplet $(a-b),\,(b-c),\,(c-a)$ must be a permutation of $(2,\,-1,\,-1)$. Without loss of generality, let

$a-b=-1$ and $c=a+2$.

This gives $b=a+1$. Substituting these values into the first equation gives

$a^2(a+1)+(a+1)^2(a+2)+(a+2)^2(a)=2186$

$a(a^2+3a+3)=2^3\cdot 7 \cdot 13$

and since $a$ is an integer, $a=8$ makes the equations true. Therefore $(a,\,b,\,c)=(8,\,9,\,10)$.

Since our WLOG will not affect the values of the triplet (but rather their order), the final answer is $a^2+b^2+c^2=8^2+9^2+10^2=245$.

 

[Albert1]

Thank you Opalg for participating and also your solution, the educated guess is spot on!

Solution of other:

Spoiler

If we subtract the first equation from the second, we get:

$ab(b-a)+bc(c-b)+ac(a-c)=2$

Now setting $a=b$ turns the LHS into $a^2(a-a)+ac(c-a)+ac(a-c)$, which after simplifying equals to 0. This means $a-b$ is a factor of the LHS.

Similarly, $b-c$ and $c-a$ are also factors. Therefore we get the factored equation $(a-b)(b-c)(c-a)=2$.

Note that $(a-b)+(b-c)+(c-a)=0$, the ordered triplet $(a-b),\,(b-c),\,(c-a)$ must be a permutation of $(2,\,-1,\,-1)$. Without loss of generality, let

$a-b=-1$ and $c=a+2$.

This gives $b=a+1$. Substituting these values into the first equation gives

$a^2(a+1)+(a+1)^2(a+2)+(a+2)^2(a+1)=2186----(1)$

$a(a^2+3a+3)=2^3\cdot 7 \cdot 13$

and since $a$ is an integer, $a=8$ makes the equations true. Therefore $(a,\,b,\,c)=(8,\,9,\,10)$.

Since our WLOG will not affect the values of the triplet (but rather their order), the final answer is $a^2+b^2+c^2=8^2+9^2+10^2=245$.

$a^2(a+1)+(a+1)^2(a+2)+(a+2)^2(a+1)=2186----(1)$
$a=8$ then $(1)=2286\neq 2186$

 

[Albert1]

$a^2(a+1)+(a+1)^2(a+2)+(a+2)^2(a+1)=2186--(1)$
if $a=8$ then $(1)=2286\neq 2186$

there must have a typo in this answer here it is: (1) sould be : $a^2(a+1)+(a+1)^2(a+2)+(a+2)^2a=2186----(1)$
sorry It seemed to be a little bit picky. it is my habit to check the answer to make sure it is correct

 

[anemone]

there must have a typo in this answer here it is: (1) sould be : $a^2(a+1)+(a+1)^2(a+2)+(a+2)^2a=2186----(1)$
sorry It seemed to be a little bit picky. it is my habit to check the answer to make sure it is correct

Hey Albert, please don't worry about pointing out the typo(s) that I've made because not only I won't upset about it, I actually appreciate your effort (and willingness) to check my solution!

Thank you Albert and I will edit my solution so that the solution is perfect now!