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

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The discussion focuses on finding the smallest multiple of 2013, denoted as k, that satisfies the given system of equations involving variables a, b, and c. The equations include a product of squared terms and a sum of sixth powers, equating to a function of k. Participants are encouraged to explore potential solutions and share their findings. The thread emphasizes collaboration and problem-solving within the mathematical community. Ultimately, the goal is to determine the least value of k that meets the criteria set by the equations.
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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.
 
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If $$a$$, $$b$$ and $$c$$ are the members of a Pythagorean triple, the system of equations has a solution.

The smallest $$abc$$ which is a multiple of 2013 is (11, 60, 61) -- minimal $$k$$ is 20 * 2013 = 40260.
 
Thanks for participating, greg1313! Your answer is correct!
 
anemone said:
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:

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

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 $v=5$.
 
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