Some turbo-charged high school algebra
Dave L. Renfro
Victor Thébault, Problem #30, Questions D'Examen column,
Mathesis Recuiel Mathématique 53 (1939), p. 203.
Here's the complete text of the problem, including
the solution sketch. The ellipses were in the original.
-------------------------------------------------------
Trouver quatre termes consécutifs d'une progression
arithmétique de raison donnée r, connaissant leur
produit a.
L'équation du problème est
(x + r)(x + 2r)(x + 3r)(x + 4r) = a.
En posant
x^2 + 5rx + 5r^2 = y,
elle se transforme en
(y - r^2)(y + r^2) = a, ...
-------------------------------------------------------
Here's an English re-statement (not a close translation)
of the problem.
PROBLEM: Given values for a and r, find a 4-term arithmetic
sequence with common difference r such that
the product of all 4 terms is equal to a.
SOLUTION SKETCH: The equation to be solved is
(x + r)(x + 2r)(x + 3r)(x + 4r) = a.
Letting y = x^2 + 5rx + 5r^2, the equation transforms to
(y - r^2)(y + r^2) = a.
-------------------------------------------------------
O-K, having just drunk a fair amount of coffee minutes
before I saw this, I decided to jump right in and verify
it (with no computer algebra system available where
I'm presently at):
(x + r)(x + 2r)
= x^2 + 3rx + 2r^2
(x + r)(x + 2r)(x + 3r)
= x^3 + 6rx^2 + 11r^2x + 6r^3
(x + r)(x + 2r)(x + 3r)(x + 4r)
= x^4 + 10rx^3 + 35r^2x^2 + 50r^3x + 24r^4
Let's see if this is equal to (y - r^2)(y + r^2)
when y = x^2 + 5rx + 5r^2:
(x^2 + 5rx + 5r^2 - r^2)(x^2 + 5rx + 5r^2 + r^2)
= (x^2 + 5rx + 4r^2)(x^2 + 5rx + 6r^2)
Expanding this out and combining like terms gives
x^4 + 10rx^3 + 35r^2x^2 + 50r^3x + 24r^4, the same
expression I got for (x + r)(x + 2r)(x + 3r)(x + 4r).
Of course, the point to the y-substitution is that
(y - r^2)(y + r^2) = a
is easily solved:
y^2 - r^4 = a
y^2 = r^4 + a
With value(s) of y in hand, then it's just
a matter of solving the quadratic equation
y = x^2 + 5rx + 5r^2
for x.
I did this, and the expression for x cleaned up a little bit,
but not sufficiently to excite me all that much, so I'll
skip the expression I got.
Of course, what I was really curious about is how anyone would
have thought to make the substitution y = x^2 + 5rx + 5r^2.
So I decided to revisit the original equation and try to
solve it myself.
(x + r)(x + 2r)(x + 3r)(x + 4r) = a
It would be nice if this was "re-centered", such as
(x - 2k)(x - k)(x + k)(x + 2k) = a,
but this is not (immediately) reducible to the original
equation, since the left hand side of the above equation
omits the term 'x' in a 5-term arithmetic sequence. Also,
I found that I needed to allow some of the change to be
accomodated by changing the variable x to another variable.
What I want is a 4-term arithmetic sequence whose terms,
relative to some new variable u, are symmetric about 0.
The simplest such sequence is probably -3, -1, 1, 3.
To get the original equation to transform to
(u - 3s)(u - s)(u + s)(u + 3s) = a,
we need u - 3s = x + r, u - s = x + 2r, etc.
These first two equations can be solved for u, s
in terms of x, r (start by subtracting the equations):
u = x + (5/2)r
s = (1/2)r
Checking to make sure everything is fine . . .
u - 3s = [x + (5/2)r] - 3(1/2)r = x + (2/2)r = x + r
u - s = [x + (5/2)r] - (1/2)r = x + (4/2)r = x + 2r
u + s = [x + (5/2)r] + (1/2)r = x + (6/2)r = x + 3r
u + 3s = [x + (5/2)r] + 3(1/2)r = x + (8/2)r = x + 4r
Thus, with these substitutions, the equation becomes
(u - 3s)(u - s)(u + s)(u + 3s) = a
(u^2 - 9s^2)(u^2 - s^2) = a
We now perform another variable change to "re-center"
the terms about 0.
Letting v^2 = u^2 - 5s^2, the equation becomes
(v^2 - 4s^2)(v^2 + 4s^2) = a
At this point it is easy to solve for v in terms
of s and a, then use this to solve for u in terms
of s and a, and finally get x in terms of r and a.
How do my variable changes relate to the single variable
change that Victor Thébault gave? Let's see . . .
I have
(v^2 - 4s^2)(v^2 + 4s^2) = a
Since s = (1/2)r, this equation is equivalent to
(v^2 - r^2)(v^2 + r^2) = a
Also, since v^2 = u^2 - 5s^2 and u = x + (5/2)r,
we have
v^2 = [x + (5/2)r]^2 - 5[(1/2)r]^2
v^2 = x^2 + 5rx + (25/4)r^2 - (5/4)r^2
v^2 = x^2 + 5rx + (20/4)r^2
v^2 = x^2 + 5rx + 5r^2
Since Thébault used y = v^2 = x^2 + 5rx + 5r^2,
my v^2 is his y, and so my equation becomes
(y - r^2)(y + r^2) = a,
which is the same as his equation.
Dave L. Renfro
Joshua Cranmer
> (x + r)(x + 2r)(x + 3r)(x + 4r)
> = x^4 + 10rx^3 + 35r^2x^2 + 50r^3x + 24r^4
Remembering some facts about algebra, the last term is 4r^2 * 6r^2... or
(5+/-1)r^2. Hmm... let's see if we can get something in the form (y -
r^2) * (y + r^2).
If this is true, we must have y = x^2 + 5r^2 + f(r, x). What other terms
need we have? An rx looks helpful. So assume f(r, x) = Arx. We then get:
(x^2 + Arx + 6r^2)(x^2 + Arx + 4r^2) =
x^4 + 2Arx^3 + (10 + A^2)r^2x^2 + 10Ar^3x + 24r^4 =
x^4 + 10rx^3 + 35r^2x^2 + 50r^3x + 24r^4. Hey, A = 5 works!
> Of course, what I was really curious about is how anyone would
> have thought to make the substitution y = x^2 + 5rx + 5r^2.
See above.
Dave L. Renfro
http://groups.google.com/group/sci.math/msg/62720599acd91fac
http://mathforum.org/kb/message.jspa?messageID=6344054
> PROBLEM: Given values for a and r, find a 4-term arithmetic
> sequence with common difference r such that
> the product of all 4 terms is equal to a.
>
> SOLUTION SKETCH: The equation to be solved is
>
> (x + r)(x + 2r)(x + 3r)(x + 4r) = a.
>
> Letting y = x^2 + 5rx + 5r^2, the equation transforms to
>
> (y - r^2)(y + r^2) = a.
Besides Joshua Cranmer's alternate method in this thread,
I got the following reply in another discussion group:
"The substitution suggested by Thebault will become obvious
upon expanding the products (x+r)(x+4r) and (x+2r)(x+3r)."
Also, I happened to come across my copy of George Chrystal's
"Algebra" text this weekend (while boxing up some of my books
and papers, for ease in moving to a new apartment or house sometime
in the near future; these were quickly put into someone's car garage
a couple of months ago when I had to evacuate due to the floods
in Iowa), and so I took the book with me to look through later,
and I managed to find a very similar problem in Volume I,
Chapter XVII, Section 7, Example 4, pp. 409-410 (this being
AMS Chelsea reprint of the 7th edition of Part I):
--------------------------------------------------------
Example 4.
(x+a)(x+a+b)(x+a+2b)(x+a+3b) = c^4.
Associating the two extreme and the two intermediate factors
on the left, we may write this equation as follows,
{x^2 + (2a+3b)x + a(a+3b)}{x^2 + (2a+3b)x + (a+b)(a+2b)} = c^4.
If eta = x^2 + (2a+3b)x + (a^2 + 3ab), the last equation may
be written
(eta)(eta + 2b^2) = c^4;
that is, (eta)^2 + 2b^2(eta) + b^4 = b^4 + c^4.
Hence eta = -b^2 +/- sqrt(b^4 + c^4).
The original equation is therefore equivalent to the
two quadratics
x^2 + (2a+3b)x + a^2 + 3ab + b^2 = +/- sqrt(b^4 + c^4).
--------------------------------------------------------
Dave L. Renfro