Separation of a Square into Two Rational Squares

Let \[ x=\frac{2zm}{1+m^2}, \qquad y=\frac{z(1-m^2)}{1+m^2}, \] where \(z\) and \(m\) are rational numbers and \(1+m^2\neq 0\). Then \[ x^2+y^2=z^2. \]

Substitute the given expressions for \(x\) and \(y\):

\[ x^2=\left(\frac{2zm}{1+m^2}\right)^2=\frac{z^2(2m)^2}{(1+m^2)^2}, \] \[ y^2=\left(\frac{z(1-m^2)}{1+m^2}\right)^2=\frac{z^2(1-m^2)^2}{(1+m^2)^2}. \]

Therefore

\[ x^2+y^2 = \frac{z^2(2m)^2}{(1+m^2)^2} + \frac{z^2(1-m^2)^2}{(1+m^2)^2}. \]

Since the denominators are equal, combine the fractions:

\[ x^2+y^2 = z^2\cdot \frac{(2m)^2+(1-m^2)^2}{(1+m^2)^2}. \]

Now expand the numerator:

\[ (2m)^2+(1-m^2)^2 = 4m^2+\left(1-2m^2+m^4\right) = 1+2m^2+m^4. \]

But

\[ 1+2m^2+m^4=(1+m^2)^2. \]

Hence

\[ x^2+y^2 = z^2\cdot \frac{(1+m^2)^2}{(1+m^2)^2} = z^2. \]

Thus

\[ x^2+y^2=z^2. \]

This proves the theorem. ∎

Bachet’s substitution works in the quadratic case because \[ (2m)^2+(1-m^2)^2=(1+m^2)^2, \] but this mechanism does not extend to powers \(n>2\). In particular, for \(n>2\), the expression \[ (2m)^n+(1-m^2)^n \] does not reduce to a single \(n\)-th power by the same algebraic pattern.

In the square case,

\[ x=\frac{2zm}{1+m^2},\qquad y=\frac{z(1-m^2)}{1+m^2} \]

gives

\[ x^2+y^2 = z^2\cdot \frac{(2m)^2+(1-m^2)^2}{(1+m^2)^2}. \]

The method succeeds because

\[ (2m)^2+(1-m^2)^2 = 4m^2+\left(1-2m^2+m^4\right) = 1+2m^2+m^4 = (1+m^2)^2. \]

Thus numerator and denominator become the same square, so cancellation yields

\[ x^2+y^2=z^2. \]

Now consider the same pattern for a higher power \(n>2\). For the method to extend in the same way, one would need the numerator

\[ (2m)^n+(1-m^2)^n \]

to recombine into a single \(n\)-th power, so that it could cancel against a denominator of the form

\[ (1+m^2)^n. \]

But for \(n>2\), binomial expansion shows that this does not occur. For example, when \(n=3\),

\[ (2m)^3+(1-m^2)^3 = 8m^3+\left(1-3m^2+3m^4-m^6\right) \] \[ (2m)^3+(1-m^2)^3 = 1-3m^2+8m^3+3m^4-m^6. \]

This is a polynomial with five distinct terms of differing degrees:

\[ 1,\quad -3m^2,\quad 8m^3,\quad 3m^4,\quad -m^6. \]

It is not a single cube obtained by the same simple quadratic cancellation pattern, and in particular it is not equal to

\[ (1+m^2)^3=1+3m^2+3m^4+m^6. \]

Likewise, when \(n=4\),

\[ (2m)^4+(1-m^2)^4 = 16m^4+\left(1-4m^2+6m^4-4m^6+m^8\right), \] \[ (2m)^4+(1-m^2)^4 = 1-4m^2+22m^4-4m^6+m^8. \]

Again this does not equal

\[ (1+m^2)^4=1+4m^2+6m^4+4m^6+m^8. \]

More generally, for every \(n>2\),

\[ (1-m^2)^n=\sum_{k=0}^{n}\binom{n}{k}(-1)^k m^{2k}, \]

so

\[ (2m)^n+(1-m^2)^n = 2^n m^n+\sum_{k=0}^{n}\binom{n}{k}(-1)^k m^{2k}. \]

This contains the term \(2^n m^n\), together with the even-degree terms

\[ 1,\ m^2,\ m^4,\ \dots,\ m^{2n}. \]

For \(n>2\), these terms do not collapse into the single binomial expansion

\[ (1+m^2)^n=\sum_{k=0}^{n}\binom{n}{k}m^{2k}, \]

because the signs and coefficients do not match, and because the extra term \(2^n m^n\) does not disappear into that pattern. Hence the numerator does not become the same \(n\)-th power as the denominator.

Therefore the special identity that makes Bachet’s method succeed is specific to the quadratic case. The cancellation mechanism is inherently quadratic and does not extend to powers greater than \(2\). ∎

From the Greek of Diophantus

To divide a given square into two squares. Let it be required to divide 16 into two squares. Let the first square be the square of some number. Then the remainder is 16 minus the square of that number, and this must be equal to a square. I form a square from whatever numbers I wish, lacking so many units as the side of 16 contains. Thus, from this construction, the remainder becomes a square. And so the given square is divided into two squares.

From the Latin of Bachet

I shall place first a square, the square of an unknown. Therefore 16 minus the square of the unknown ought to be a square. I feign a square from a root, if one pleases, 2q − 4. And so 16 − q² will be equal to 4q² − 16q + 16. Therefore 5q² are equal to 16q, and q becomes 16/5. Accordingly the first square is 256/25, and the remaining one is 144/25.

From the Latin of Fermat, written beside Problem 8

It is impossible to separate a cube into two cubes, or a fourth power into two fourth powers, or in general any power greater than the second into two powers of the same kind. I have discovered a truly remarkable demonstration of this, which this margin is too small to contain.