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.

Some time ago I saw that the record for Broccard's Problem ( http://en.wikipedia.org/wiki/Brocard%27s_problem ) was pretty low($10^{9}$) because of the scarce attention it has, so I coded a program to prove it up to $10^{11}$. Then I presented it at my school's mathfest. The other presentations were just informative about popular things like fibonacci or pascal, and it really surprised the jury :). I don't think it was a big deal, after all neither my code or my computer were too fast, and it only took 2 days. Any decent investigator can easily do way better than that in a couple of days.
But it was a very nice experience discovering something new for the first time on my own.
