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anemone
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Prove that the equation $x_1^3+x_2^3+x_3^3+x_4^3+x_5^3=k$ has an integer solution for any integer $k$.
magneto said:The hint follows from:
\[
(x+1)^3 + (x-1)^3 + (-x)^3 + (-x)^3 = 6x.
\]
Moroever, we have that under $\pmod 6$,
$(\pm 1)^3 \equiv \pm1$, $(\pm 2)^3 \equiv \pm 2$,
$3^3 \equiv 3$. We can choose any of the two of the cubes to form
$6x + k$ where $k = 1..5$.
To prove that there exists an integer solution for this equation, we can use the concept of modular arithmetic. We can show that for any integer $k$, there exists a set of 5 integers $(x_1, x_2, x_3, x_4, x_5)$ that satisfy the equation by considering all possible remainders when each of the integers is divided by 9. We can then use the properties of modular arithmetic to manipulate the equation and show that it is always possible to find a set of integers that satisfy it.
Yes, an example of an integer solution for this equation is $(x_1, x_2, x_3, x_4, x_5) = (1, 2, 3, 4, 5)$ for $k = 225$. This set of integers satisfies the equation $1^3+2^3+3^3+4^3+5^3=225$.
The number 9 is significant because it is the base of the decimal system, and thus has special properties in modular arithmetic. In particular, when we take the remainder of a number divided by 9, we are essentially looking at the digit sum of that number. This allows us to manipulate the equation in a way that helps us find integer solutions.
Yes, there are other methods such as using the properties of complex numbers or the properties of perfect cubes. However, the method involving modular arithmetic is the most commonly used and efficient approach.
Yes, this equation can have multiple sets of integer solutions for a given value of $k$. For example, for $k=225$, in addition to the solution (1, 2, 3, 4, 5), there is also the solution (6, 6, 6, 6, 0) and many others. This can be shown using the properties of modular arithmetic and the fact that the equation has infinitely many solutions for any integer $k$.