This question is inspired by
my answer to this one:
Surprising identities / equations
In that question, people were asked about
the most surprising result
that they knew.
Almost all of them quoted someone
else's result.
I was one of the only ones
to reply about a result of mine
that greatly surprised me.
So, I have decided to make that
a question on its own:
What is your own mathematical result
that surprised you the most?
Here is mine.
Consider the diophantine equation
$$x(x+1)…(x+n-1) -y^n = k$$
where $x, y, n,$ and $k$ are integers,
$x \ge 1$,
$y \ge 1$,
and
$n \ge 3$.
I was led to consider considering this
by trying to generalize the
Erdos-Selfridge result
that the product of consecutive integers
could never be a power.
I phrased this as
"How close and how often
can the product of
$n$ consecutive integers
be to an $n$-th power?"
Looking at this equation,
it seemed reasonable to think that,
for fixed $k$ and $n$,
there were only a finite number of
$x$ and $y$ that satisfied it.
This was not too hard to prove.
What greatly surprised me was that
I was able to prove that
for any fixed $k$,
there were only a finite number of
$n$, $x$, and $y$ that satisfied it.
The proof went like this:
I first showed that
any solution must have
$y \le |k|$.
This was moderately straightforward,
and involved considering the three cases
$y < x$, $x \le y \le x+n-1$,
and $y \ge x+n$.
Note: The proof that
$y \le |k|$
has been added at the end.
The next step really surprised me.
I showed that
$n < e|k|$,
where $e$ is the good old base of natural logarithms.
The proof was amazingly (to me) simple.
Since $y \le |k|$
and
$2(n/e)^n < n!$,
$\begin{align}
2(n/e)^n
&< n!\\
&\le x(x+1)…(x+n-1)\\
&= y^n+k\\
&\le |k|^n+|k|\\
&\le |k|^n+|k|^n\\
&= 2|k|^n\\
\end{align}
$
so
$n < e |k|$.
I still remember staring at this
in disbelief,
over forty years later.
I was asked to show my proof
that $y \le |k|$.
For brevity,
I will write
$x(x+1)…(x+n-1)$
as $x!n$,
because this is a generalization of
factorial.
The basic inequality is
$$(x^2+(n-1)x)^{n/2} \le x!n \le (x+(n-1)/2)^n$$
I also use two lemmas:
(L1) If $0 < a < b$ and $n > 1$
then
$n(b-a)a^{n-1} < b^n-a^n < n(b-a)b^{n-1}$.
(L2) If $a^m \leq b^m+c$ where
$a \geq 0$,
$b >0$, $c \geq 0$,
and $m \geq 1$,
then
$a \leq b + c/(m\,b^{m-1})$.
The basic idea is simple:
either $x < y < x+n-1$ or $y$ is outside this range.
If $y$ is inside the range,
then $y$ divides both $x!n$ and $y^n$,
so $y$ divides their difference,
which is $k$.
If $y$ is outside the range,
then we can use
the basic inequality
and the lemmas
to derive very strong inequalities
on $x$ and $y$.
Here are all the cases.
If $k=0$, so
$x!n = y^n$,
then $x < y < x+n-1$, or
$x+1 < y+1 \leq x+n-1$,
so that $y+1 | x!n$ or
$y+1 | y^n$, which is impossible.
If $k > 0$,
$x!n > y^n$, so that,
$y < x+(n-1)/2$.
If $ y > x$, then,
as stated above, $y | k$.
If $y \leq x$, then
$(x^2 + (n-1)x)^{n/2} \le x!n
= y^n + k
\le x^n + k
$
or, by L2,
$x^2 + (n-1)x \le
x^2 + 2k/\left(n\,x^{n-2}\right)
$
so that
$ x^{n-1} \leq 2k/n(n-1). $
Therefore
$y \le x \le \left(\frac{2k}{n(n-1)}\right)^{1/(n-1)}$.
If $k < 0$,
$x!n < y^n$, so that,
$y^2 > x^2+(n-1)x$,
which implies that $y > x$.
If $ y < x+n-1$, then,
as stated above, $y | |k|$.
If $y \geq x+n-1$, then
$(x+n-1)^n \leq y^n
= x!n – k
= x!n + |k|
\leq (x+(n-1)/2)^n + |k|
$
or, by L2,
$x+n-1 \leq
x+(n-1)/2 +
\frac{|k|}{
n(x+(n-1)/2)^{n-1}
}$
or
$(n-1)/2
\leq \frac{|k|}
{ n(x+(n-1)/2)^{n-1}} $
so that
$\left(x+(n-1)/2\right)^{n-1} \leq \frac{2|k|}{n(n-1)}.
$
Since
$y^n \leq (x + (n-1)/2))^n + |k|
\leq \left(\frac{2|k|}{n(n-1)}\right)^{n/(n-1)} + |k|
\leq |k|^{n/(n-1)} + |k|,
$
$ y \leq |k|^{1/(n-1)} + 1/n.$
In all the cases,
$y \le |k|$.
When $y < x$ or $y \ge x+n-1$,
$y$ is significantly smaller.
Best Answer
I was very happy to find out that if we look at a notebook with a magnifying glass, then the lines become curves; and the fact that they are parallel is remained (especially if you keep them at the focal point of the magnification).
However the curves all meet at the "edge" of the glass. So we can have a sense of geometry where parallel lines meet at infinity.
I remember telling about this to my brother who was an engineering student (I was merely 16), and he said that it's impossible. Some years later I learned that this was already known as non-Euclidean geometry and played an important part of Einstein's relativity.