March 2005 solution

The answer is x + y = 0. There were many interesting approaches to the solution, and Hecquet single-handedly provided a version of each. We shall use his rubrics together with a combination of nice ideas from each contributor. Most of the solutions make use of an easily verified fact:

1. A la hussarde (The method of Agarwal, de la Viuda, Kais, LaLiberté, Phan, and Wood). The idea here is to get rid of the radicals by squaring, but even a hussar would simplify things a bit before squaring. Among the various possibilities for simplification, we can use our basic fact to rewrite the given equation as

Square both sides

then subtract 1 from both sides to get,

Since the terms in parentheses are equal and nonzero, we can cancel them and divide by 2 to get y = –x, so that x + y = 0, as claimed. Of course, after we square the sides of an equation we must check that no new solutions have been introduced: one checks easily that y = –x satisfies the given equation.

2. Un peu plus fûté. (The method of Amghibech, Barroso, and Moutaharir.)

This time let's use our basic fact twice (once with x and once with y):

Add these two equalities to find x + y + R = –(x + y) + R, where R is the sum of two radicals; thus, 2(x + y) = 0, which again reduces to x + y = 0.

3. A la Euler. (The method of Cahangirov, Lo Presti, Loyer, Anonymous, and Tran.)

Let Then so that k² -2x + x² = x² - 1 from which

The given equation implies that which (by using the same steps as before) leads to

Adding these two expressions, we find that x + y = 0.

Comment. Those who are familiar with hyperbolic functions will recognize what is going on here. The hyperbolic functions are related to the hyperbola x² – y² = 1 in a manner analogous to the relation of the trigonometric functions to the circle x² + y² = 1. (Actually, the hyperbolic functions were not introduced by Euler, but by his contemporary Vicenzo Riccati (1707-1775), so maybe we should have entitled method 3, "A la Riccati.") The hyperbolic sine

(usually pronounced "sinch") and the hyperbolic cosine

satisfy the basic relation

cosh² a - sinh² a = 1.

Other easily verified identities that we use here are cosh a + sinh a = e^a and sinh(–a) = –sinh a. Note that cosh a is strictly positive (that is, ); it takes every value greater than 1 exactly twice. On the other hand, sinh a is strictly increasing, taking all real values from –∞ to ∞ once and only once; in particular, there is a u and a v for which sinh u = x and sinh v = y. If it is not yet obvious, here is their relationship to the previous argument: define u = ln k; that is, e^u = k. The above equation for x then becomes

So, here is the argument once again in the revised notation. The given equations become

This last equation implies that sinh u = sinh(–v) = –sinh v, which says that x = –y, so that x + y = 0.

4. Euler bis. (The method of Carpentier and Lichtenberg.)

Here we use the inverse hyperbolic functions: x = sinh u if and only if u = sinh^-1x. It is easily verified that

This observation leads to the equivalent equations

5. Appel à l'âne à Lise. (Bornsztein's method.)

The argument above in (4) is easily seen to apply in a more general setting. We start by setting . We note that that is,

f(x) > 0.

Furthermore,

x < 0 if and only if f(x) < 1,

since the latter inequality is equivalent to which (by squaring the equal positive numbers) says simply that 0 < -2x. On the other hand, from the definition of f(x),

x > 0 if and only if f(x) > 1.

Consequently, if f(x)f(y) = 1 we can assume without loss of generality that x ≤ 0 ≤ y. Our basic fact says that f(x) satisfies 1/f(x) = f(–x), so our given equation is

f(y) = f(-x).

Because f(x) is the sum of two functions that are strictly increasing as x runs from 0 to infinity, f(x) itself is strictly increasing. Since y ≥ 0 and –x > 0, we deduce that y = –x and, therefore, x + y = 0.

6. A la Euclid. This approach was unique to Hecquet. He presented only the start of a pictorial argument. We modified it a bit and (in the words of people who don't quite know how to finish the argument) we leave the details to the reader. Our picture is based on two Euclidean theorems:

If a secant from an external point P meets a circle at the points Q and R, and one tangent from P touches that circle at T, then PQ x PR = PT2.

and

The radius from the center C to a point T on the circle is perpendicular to the tangent at T.

The accompanying figure shows the right triangle PTC with legs x and 1 and hypotenuse so that PQ x PR =