How to show that there exists infinitely many triples $(x,y,z)$ of integers such that ...
问题内容
This question was asked to me by a junior trying number theory problems and I got struck on it.
Question:Show that there are infinitely many triples $(x,y,z)$ of integers such that $x^3+y^4=z^{31}$.
I found a solution here https://www.isical.ac.in/~rmo/papers/rmo/rmo-2015-3.pdf : as problem $3$ in CRMO while browsing MSE.
Solution given on the website whose link doesn't work when I post the question but works when I copy the link and paste it in browser: Choose $x = 2^{4r}$ and $y = 2^{3r}$.Then the left side is $2^{12r+1}$. If we take $z = 2^k$, then we get $2^{12r+1} = 2^{31k}$ .Thus it is sufficient to prove that the equation $12r + 1 = 31k$ has infinitely many solutions in integers. Observe that $(12 × 18) + 1 = 31 × 7.$ If we choose $r = 31l + 18$ and $k = 12l + 7,$ we get $12(31l + 18) + 1 = 31(12l + 7),$ for all $l.$ Choosing $l \in \mathbb{N}$, we get infinitely many $r = 31l + 18$ and $k = 12l + 7$ such that $12r + 1 = 31k.$ Going back we have infinitely many $(x, y, z)$ of integers satisfying the given equation.
But the solution given here seems very non -intuitive to me. How I am supposed to know that I should choose $x=2^{4r}$ and $y=2^{3r}$ and overall the solution looks quite non-elegant to me.
Question: I have studied number theory from David Burton Elementary Number Theory and Tom M Apostol Introduction to Analytic Number Theory. Is there an elegant alternative solution which uses some advanced method given in these textbooks?
Can you please outline an alternative solution?
I shall be very grateful.
回答 (1)
(Insufficient space in the comments) Your question's conditions are too lenient. If you only require "integers" instead of "positive integers," the answer is actually obvious. For example, let $y = 0$, then $x^3 = z^{31}$. Taking $x = a^{31}$ and $z = a^3$ gives infinitely many solutions.
Next, you can also use the Chinese Remainder Theorem:
Let the solution be in the form of $x = 2^a, y = 2^b, z = 2^c$. We hope that:
$$(2^a)^3 + (2^b)^4 = 2^{3a} + 2^{4b} = 2^{31c}$$
If we let $3a = 4b = k$, the expression becomes:
$$2^k + 2^k = 2 \cdot 2^k = 2^{k+1}$$
Therefore, we only need that $k$ is a multiple of $3$, $k$ is a multiple of $4$, and $k+1$ is a multiple of $31$. This is equivalent to solving a system of congruence equations:
$$\begin{cases} k \equiv 0 \pmod{12} \\ k \equiv 30 \pmod{31} \end{cases}$$
Since $\gcd(12, 31) = 1$, by the Chinese Remainder Theorem, there must exist infinitely many integer solutions $k$
$$Q.E.D.$$
Additional Information: This explicitly parameterizes an infinite number of positive solutions:
$x = 2^{124m + 72}$
$y = 2^{93m + 54}$
$z = 2^{12m + 7}$