What Is the Smallest Multiple of 2013 That Solves This System of Equations?

[anemone]

Find the least multiple $k$ of $2013$ for which the system of equations

$(a^2+b^2)(b^2+c^2)(c^2+a^2)=a^6+b^6+c^6+4k^2$

$abc=k$

has a solution.

 

[Greg]

Spoiler

If \(\displaystyle a\), \(\displaystyle b\) and \(\displaystyle c\) are the members of a Pythagorean triple, the system of equations has a solution.

The smallest \(\displaystyle abc\) which is a multiple of 2013 is (11, 60, 61) -- minimal \(\displaystyle k\) is 20 * 2013 = 40260.

 

[anemone]

Thanks for participating, greg1313! Your answer is correct!

 

[anemone]

Find the least multiple $k$ of $2013$ for which the system of equations

$(a^2+b^2)(b^2+c^2)(c^2+a^2)=a^6+b^6+c^6+4k^2$

$abc=k$

has a solution.

Solution of other:

Spoiler

$a^6+b^6+c^6+4(abc)^2-(a^2+b^2)(b^2+c^2)(c^2+a^2)=(a^2-b^2-c^2)(a^2+b^2-c^2)(a^2-b^2+c^2)$

Therefore we get:

$(a^2-b^2-c^2)(a^2+b^2-c^2)(a^2-b^2+c^2)=0$

And deduce that either $a^2-b^2-c^2=0$ or $a^2+b^2-c^2=0$ or $a^2-b^2+c^2=0$.

This means $(a,\,b,\,c)$ is a Pythagorean triple and is therefore of the form $(u^2-v^2,\,2uv,\,u^2+v^2)$ for certain integers $u$ and $v$.

So,

$abc=k=n(2013)=n(3)(11)(61)=(u^2-v^2)(2uv)(u^2+v^2)=(2uv)(u-v)(u+v)(u^2+v^2)$

One easily verifies that the least multiple of $2013$ for which the above equation holds is $k=20(2013)=40260$, with $u=6$ and $v=5$.