Prove that the equation $x^3 + 10000 = y^3$ has no solutions

Question

The below question was given in my workbook for number theory. I have seen the solution, and it utilizes $\mod 7$, but I am unsure of why $\mod 7$ was chosen to solve the problem. Would any number of alternatives suited the problem?

Prove that the equation $x^3 + 10000 = y^3$ has no solutions.

Answer 1

I think the main question is "Why choose $\pmod 7$?"

Well, first of all, $7 \nmid 10000$ -- in particular, $10000 \equiv 4 \pmod 7$. If we looked at $\pmod 5$, it would be useless because we can have $x, y$ be multiples of $5$ and our method would not work.

Second of all, $7 = 2^3 - 1$. Thus, we have $(2n)^3 \equiv n^3 \pmod 7$ and for $n \in \{1, 2, 3\}$ the residue is either $-1$ or $1$. Thus, we can also say for odd numbers that $(7 - 2n)^3 \equiv (-2n)^3 \equiv -n^3 \pmod 7$, so that's an explanation of why the solution chose $7$ (it can only take on residues of $-1, 0, 1 \pmod 7$)

Now, it is clear that adding $-1, 0,$ or $1$ to $4$ can never result in $-1, 0,$ or $1 \pmod 7$

Answer 2

An integer cube will always leave a remainder of either $0, 1, 6$ when divided by $7$ (I suppose that I will leave this to you to prove), while $10000$ leaves a remainder of $4$ when divided by $7$.

So we see that if we divide $x^3 + 1000$ by $7$, it must leave a remainder of either $4, 5, 3$ (by summing up both of their remainders in the ring of integers modulo $7$).

On the other hand, $y^3$ is an integral cube. As disused earlier, all integer cubes must leave a remainder of either $0, 1, 6$ when divided by $7$.

Since the remainder of both sides when divided by $7$ will never be equal, there are no integer solutions.

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A possible alternative would be to rearrange the given problem into:

$$y^3 - x^3 = 10000$$

Now, we factorize this to get:

$$(y-x)(y^2 + xy + x^2) = 10000$$

Over here, we can test all possible factors of $10000$ to see that no such integral $x, y$ exists.

Answer 3

$7$ is a good number to try because the multiplicative group of non-zero residues modulo a prime $p$ is cyclic of order $p-1$. [Order $p-1$ is very easy, cyclic takes a bit more work, note $7-1=6$ is divisible by $3$]. This means that the cubes modulo $7$ will be $0$ and $\frac {7-1}3=2$ other numbers which turn out to be $\pm1$. This is a particularly simple set of things to be working with, and it is easy to see that $$\{0,\pm1\}+4=\{3,4,5\}$$

If this hadn't worked, the primes to try are the ones of the form $6n+1$ - the cubes mod $13$ are $0,\pm 1, \pm 8$ - there are $\frac {13-1}3=4$ non-zero cubes modulo $13$. $10000$ leaves remainder $3$. $$\{0,\pm1,\pm 8\}+3=\{3,2,4,11,-5\}$$ but modulo $3$ we have $-5=8$ and we have more work to do. As numbers increase, we have more cases to consider.

If we choose a prime $p=6n-1$ we find that every element is a cube (the cyclic group of order $6n-2$ has no elements of order $3$, which would cube to $1$), and this approach gets us nowhere.