Given the number (c), and told that it equals (a)*(b),
where all
three are different odd numbers, then when attempting
to
develop an algorithm to find (a) and (b), one of the very
common things that appears in the mathematics is the
Quadratic Formula. To refresh your memory about that,
see
the first link.

An example line of reasoning, that leads to needing the
quadratic formula, goes like this:

I know that (a)*(b) = (c)

Suppose I knew what (a)+(b) added to equal? Call it (d),
to
represent it, even though I don't actually know it.

If (a)+(b)=(d), then it logically follows that (b) = (d)-(a)

I can plug that into the multiplication equation, like this:

(a)*[(d)-(a)] = (c)

That can be algebraically manipulated to become this:

[(a)*(d)] - [(a)*(a)] = c --and then:

0 = c + [(a)*(a)] - [(d)*(a)] --and then:

[(a)*(a)] - [(d)*(a)] + c = 0

That last equation is suitable for applying the quadratic
formula. Please don't be confused by the way certain
variables
are used here, and the way they are used in the formula.
There, the (a) and (b) and (c) of the formula are "coefficients"
of a descending series of "powers" of (x) --including x-raised-to-
the-zero-power (an invisible 1 that (c) multiplies). Here I'm
using (a) instead of (x), and the relevant/respective
coefficients of powers of (a) are an invisible multiplying (1) for
the second power of (a), a (-d) for the first power of (a), and
(c) for the zeroth power of (a). Thus we will end up with this:

(a) = [(d) ± sqrt([(d)*(d)] - [(4)*(c)])] / 2

Now we could make guesses regarding (d), and see what
value
of (a) is computed --which should be one of the factors of
(c).
Indeed, because of that ± symbol (add or subtract), the
quadratic formula can always compute two results, and in
this
case both results will be factors of (c).

Feel free to experiment with some known values. For
example,
since we know that 5*7=35, we could say (a)=5, (b)=7,
(c)=35,
and (d)=5+7=12. So plug the 12 and the 35 into the
formula,
and see it produce 5 and 7.

Now here's the thing. During a number of years of playing
with
algebra and the problem of finding the factors of (c),
every
time I found an application of the quadratic formula to
compute
one of the factors of (c), the formula would also always
be able
to compute the other factor of (c).

Until just a day or two ago, as I write this. That's the
Quadratic Mystery! Why in this new formula (below) did it
only
compute one of the factors of (c)? I think the answer has
something to do with "rigging" the logic....

Here's the initial line of reasoning:

If I assume that (b) is larger than (a) --see the start of this
text, where it states that (a), (b), and (c) are all different
from each other-- then I could imagine creating an
equation to
compute (b) from (a), like this:

If (a)=5 and (b)=13, then (a)+8 = 13= (b), or [(a)*2] + 3 =
13 =
(b)

I chose to use the second possibility, because very often
(b) is
greater than (a) by several times, and I didn't want the
number
I added (8 or 3 above) to get really large --it will never
need to
be
larger than (a) if I use the multiplier version. So, I need
two
variables, one for the unknown multiplier --call it (d)--
and one
for the unknown amount to add --call it (e). Therefore:

[(a)*(d)] + (e) = (b)

I can plug THAT into the original equation (a)*(b) = (c),
and
get:

(a)*{[(a)*(d)] + (e)} = (c) --which can become:

[(a)*(a)*(d)] + [(a)*(e)] = (c) --which can become:

[(d)*(a)*(a)] + [(e)*(a)] - (c) = 0 --which is now ready for
the
quadratic formula. We still want to compute (a), but now
the
relevant/respective coefficients of the powers of (a) are (d),
(e), and (-c):

(a) = [(-e) ± sqrt([(e)*(e)] + [4*(d)*(c)])] / [2*(d)]

When known values are plugged into the formula to test
it, it
can compute two different values for (a) --one will be a
positive number and the other will be a negative number-
- and
when either of those values is plugged into the original
equation:

(a)*{[(a)*(d)] + (e)} = (c)

both result in a valid calculation of (c). But only one of
them is
actually a factor of (c)! That value is the positive number
(replace the ± symbol with a simple +), which also is (a),
the
**smaller** factor of (c). (But then see above about
"rigging"
the logic....)

Now it just so happens that a slight variation of the
formula is
able to compute the other factor of (c), which is (b):

(b) = [(e) + sqrt([(e)*(e)] + [4*(d)*(c)])] / 2

You don't have to take my word for it; you can take the
equation

(a)*(b)=(c)

and replace both (a) and (b) with the complicated
formulas, and it works --a mess of algebra can ultimately
yield
(c)=(c) --which technically counts as proof the
second version of the quadratic formula is as valid as the
first (with the ± symbol replaced by +),
each for computing ONE factor of (c).

Is all this mathematical playing around actually useful for
finding the factors
of (c) when it is a really big number (like a thousand digits
long)? I think the answer to that is "NO!". But I also
thought
this particular unusual thing was interesting
enough to post here at the HalfBakery. Hope you liked it!