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.

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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, ...

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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.

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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.