What is the most surprising result that you have personally discovered? 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.

Well, "a long time ago" (1970s) it was not so clear that integrating restrictions of Eisenstein series on big groups against cuspforms on smaller reductive groups, or oppositely, etc., would do anything interesting... much less produce $L$-functions. The Rankin-Selberg example from 1939 was not necessarily clearly advocating thinking in such terms, and Langlands' 1967/76 observations that constant terms of Eisenstein series involved $L$-functions was also easy-enough to rationalize as a thing-in-itself. So when I noticed (out of idle curiosity, being aware of somewhat-vaguer results of H. Klingen from 1962 and a general pattern of qualitative results of G. Shimura in the early 1970s) that various Euler products arose in this way (e.g., by $Sp(m)\times Sp(n)\to Sp(m+n)$), and triple-product $L$-functions (from the related $SL_2\times SL_2\times SL_2\to Sp_3$), it was quite a surprise to me. There were no similar results at the time, so there was nothing to compare to.
