Problems dependent on a parameter whose value makes every variation a new safari.

Question

I am looking for problems dependent on a parameter whose value makes every variation a new safari.

A good example of this would be the equation $x^n + y^n = z^n$ for integers $n, x, y, z$ with $xyz \ne 0.$ The case $n = 0$ has no solutions because you get $1+1=1.$ The case $n=1$ just corresponds to addition, which we know happens all the time. The case $n = 2$ leads to the theory of Pythagorean triples and is settled by the parametrization $\{x, y\} = \{2mn, m^2-n^2\}, z = m^2+n^2.$ The case $n = 4$ uses a nice argument by infinite descent, the case $n = 3$ uses a sequence of careful deductions in elementary number theory after a lemma about integers $s$ satisfying $s^3 = u^2+3v^2.$ The case $n = 5$ was done by Dirichlet and Legendre involving some casework and techniques which are different enough from Euler's proof for $n = 3.$

The case $n = -1$ is resolved by a parametrization which can be derived (and proved to give all solutions in the process) in a different way than the case $n = 2.$ For $n = k < -2,$ the non-existence of solutions is equivalent to that of $n = -k.$ Now we can consider $n$ to be a regular prime, and $n$ to be any integer $>2$ in general, so all the adventure eventually comes to an end. However, in the process we have found 6 or 7 different values of $n$ that all lead to wildly different adventures. Never have I seen before such a problem where varying the parameter can give you so many different scenarios with so many different approaches, so I'm wondering if there are any more problems like this.

Answer 1

Let $p,q,r,\dots$ be distinct primes. The number of (isomorphism classes of) groups of order $p$ is one. The number of groups of order $pq$ can be one or two. The number of groups of order $pqr$ can be one (e.g., $3\cdot5\cdot17$) or two (e.g., $3\cdot5\cdot7$) or three (e.g., $3\cdot7\cdot29$) or four (e.g., $2\cdot3\cdot5$) or five (e.g., $3\times7\times13$ or six (e.g., $2\cdot3\cdot7$) or seven (e.g., $3\cdot7\cdot43$) - in fact, it can be $1,2,3$ or any number of the form $n+2$ or $n+4$ with $n$ prime.

I don't know what happens with four or more distinct primes.

J. Alonso, Groups of square-free order, an algorithm, Math. Comp. 30 (1976), 632-637, https://doi.org/10.1090/S0025-5718-1976-0506898-5 is a nice reference.

Answer 2

The Mordell equation is $y^2=x^3+k$. The problem is, for a given integer $k$, to find all the solutions in integers $x,y$, or to prove there are none. The methods used vary widely from one value of $k$ to another. I strongly recommend Keith Conrad's survey, Examples of Mordell's equation, available at https://kconrad.math.uconn.edu/blurbs/gradnumthy/mordelleqn1.pdf

Some examples, taken from that survey:

$k=7$, proved to have no solutions, key idea being that $-1$ is a quadratic residue modulo an odd prime $p$ if and only if $p\equiv1\bmod4$. Other examples depend on the primes for which $2$ and/or $-2$ is a quadratic residue.

$k=16$, a proof that the only solutions are $(x,y)=(0,\pm4)$ uses the Unique Factorization Theorem. Other examples use unique factorization in ${\bf Z}[\sqrt{-1}]$ or ${\bf Z}[\sqrt2]$.

The case $k=1$ has a proof that relies on $p$-adic analysis.

The case $k=-26$ leads to calculations in ${\bf Z}[\sqrt{-26}]$, which is not a unique factorization domain, so more advanced techniques of Algebraic Number Theory are needed (but the paper stops short of providing these).

In a series of slides at https://kconrad.math.uconn.edu/ross2008/mordell.pdf Conrad repeats some of the examples from the paper, and then ties the Mordell equation to bounds on solutions found by Baker and by Stark, a bound conjectured by Hall, and the $abc$ conjecture.