Commentary on Proposition X.28

Back to Proposition X.28.

Euclid takes three straight lines commensurable in square only, i.e. of the form $\rho$, $k^{\frac{1}{2}}\rho$, $\lambda^{\frac{1}{2}}\rho$, and proceeds as follows. Take the mean proportional to $\rho$, $k^{\frac{1}{2}}\rho$, i.e. $k^{\frac{1}{4}}\rho$. Then take $x$ such that

$$k^{\frac{1}{2}}\rho : \lambda^{\frac{1}{2}}\rho = k^{\frac{1}{4}}\rho : x \ \text{.........................................(1),}$$

so that $x = \lambda^{\frac{1}{2}}\rho / k^{\frac{1}{4}}$. $k^{\frac{1}{4}}\rho$, $\lambda^{\frac{1}{2}}\rho / k^{\frac{1}{4}}$ are the required medial straight lines. For $k^{\frac{1}{4}}\rho$ is medial. Now, by (1), since $k^{\frac{1}{2}}\rho \sim \lambda^{\frac{1}{2}}\rho$,

$$k^{\frac{1}{4}}\rho \sim x,$$

whence $x$ is also medial [X. 23, addition], while $\sim k^{\frac{1}{4}}\rho$. Next, by (1),

$$\lambda^{\frac{1}{2}}\rho : x = k^{\frac{1}{2}}\rho : k^{\frac{1}{4}}\rho$$

$$= k^{\frac{1}{4}}\rho : \rho,$$

whence

$$x \,.\, k^{\frac{1}{4}}\rho = \lambda^{\frac{1}{2}}\rho^2\text{, which is medial.}$$

The straight lines $k^{\frac{1}{4}}\rho$, $\lambda^{\frac{1}{2}}\rho / k^{\frac{1}{4}}$ of course take different forms according as the original straight lines are of the forms (1) $a$, $\sqrt{B}$, $\sqrt{C}$, (2) $\sqrt{A}$, $\sqrt{B}$, $\sqrt{C}$,

  1. $\sqrt{A}$, $b$, $\sqrt{C}$, and (4) $\sqrt{A}$, $\sqrt{B}$, $c$.

E.g. in case (1) they are $\sqrt{a\sqrt{B}}$, $\sqrt{\frac{aC}{\sqrt{B}}}$, in case (2) they are $\sqrt[4]{AB}$, $\sqrt{\frac{C\sqrt{A}}{\sqrt{B}}}$, and so on.

LEMMA I.

To find two square numbers such that their sum is also square.

Let two numbers AB, BC be set out, and let them be either both even or both odd.

Then since, whether an even number is subtracted from an even number, or an odd number from an odd number, the remainder is even, [IX. 24, 26]
therefore the remainder AC is even.
A D C B

Let AC be bisected at D.

Let AB, BC also be either similar plane numbers, or square numbers, which are themselves also similar plane numbers.

Now the product of AB, BC together with the square on CD is equal to the square on BD. [II. 6]
And the product of AB, BC is square, inasmuch as it was proved that, if two similar plane numbers by multiplying one another make some number the product is square. [IX. 1]
Therefore two square numbers, the product of AB, BC, and the square on CD, have been found which, when added together, make the square on BD.
And it is manifest that two square numbers, the square on BD and the square on CD, have again been found such that their difference, the product of AB, BC, is a square, whenever AB, BC are similar plane numbers.
But when they are not similar plane numbers, two square numbers, the square on BD and the square on DC, have been found such that their difference, the product of AB, BC, is not square.
Q. E. D.

Euclid’s method of forming right-angled triangles in integral numbers, already alluded to in the note on I. 47, is as follows.

Take two similar plane numbers, e.g. $mnp^2$, $mnq^2$, which are either both even or both odd, so that their difference is divisible by 2.

Now the product of the two numbers, or $m^2n^2p^2q^2$, is square, [IX. 1] and, by II. 6,

$$mnp^2 \cdot mnq^2 + \left(\frac{mnp^2 - mnq^2}{2}\right)^2 = \left(\frac{mnp^2 + mnq^2}{2}\right)^2,$$

so that the numbers $mnpq$, $\frac{1}{2}(mnp^2 - mnq^2)$ satisfy the condition that the sum of their squares is also a square number.

It is also clear that $\frac{1}{2}(mnp^2 + mnq^2)$, $mnpq$ are numbers such that the difference of their squares is also square.

Lemma 2.

To find two square numbers such that their sum is not square.

For let the product of AB, BC, as we said, be square, and CA even, and let CA be bisected by D.

A G H D E F C B
It is then manifest that the square product of AB, BC together with the square on CD is equal to the square on BD. [See Lemma 1]

Let the unit DE be subtracted;

therefore the product of AB, BC together with the square on CE is less than the square on BD.

I say then that the square product of AB, BC together with the square on CE will not be square.

For, if it is square, it is either equal to the square on BE, or less than the square on BE, but cannot any more be greater, lest the unit be divided.

First, if possible, let the product of AB, BC together with the square on CE be equal to the square on BE, and let GA be double of the unit DE.

Since then the whole AC is double of the whole CD, and in them AG is double of DE,
therefore the remainder GC is also double of the remainder EC; therefore GC is bisected by E.
Therefore the product of GB, BC together with the square on CE is equal to the square on BE. [II. 6]
But the product of AB, BC together with the square on CE is also, by hypothesis, equal to the square on BE;
therefore the product of GB, BC together with the square on CE is equal to the product of AB, BC together with the square on CE.
And, if the common square on CE be subtracted,
it follows that AB is equal to GB: which is absurd.
Therefore the product of AB, BC together with the square on CE is not equal to the square on BE.

I say next that neither is it less than the square on BE.

For, if possible, let it be equal to the square on BF, and let HA be double of DF.

Now it will again follow that HC is double of CF; so that CH has also been bisected at F, and for this reason the product of HB, BC together with the square on FC is equal to the square on BF. [II. 6]
But, by hypothesis, the product of AB, BC together with the square on CE is also equal to the square on BF.
Thus the product of HB, BC together with the square on CF will also be equal to the product of AB, BC together with the square on CE: which is absurd.
Therefore the product of AB, BC together with the square on CE is not less than the square on BE.
And it was proved that neither is it equal to the square on BE.
Therefore the product of AB, BC together with the square on CE is not square. Q. E. D.

We can, of course, write the identity in the note on Lemma 1 above (p. 64) in the simpler form

$$mp^2 \cdot mq^2 + \left(\frac{mp^2 - mq^2}{2}\right)^2 = \left(\frac{mp^2 + mq^2}{2}\right)^2,$$

where, as before, $mp^2$, $mq^2$ are both odd or both even.

Now, says Euclid,

$$mp^2 \cdot mq^2 + \left(\frac{mp^2 - mq^2}{2} - 1\right)^2 \text{ is not a square number.}$$

This is proved by reductio ad absurdum.

The number is clearly less than $mp^2 \cdot mq^2 + \left(\frac{mp^2 - mq^2}{2}\right)^2$, i.e. less than $\left(\frac{mp^2 + mq^2}{2}\right)^2$.

If then the number is square, its side must be greater than, equal to, or less than $\left(\frac{mp^2 + mq^2}{2} - 1\right)$, the number next less than $\frac{mp^2 + mq^2}{2}$.

But (1) the side cannot be $> \left(\frac{mp^2 + mq^2}{2} - 1\right)$ without being equal to $\frac{mp^2 + mq^2}{2}$, since they are consecutive numbers.

  1. $(mp^2 - 2) mq^2 + \left(\frac{mp^2 - mq^2}{2} - 1\right)^2 = \left(\frac{mp^2 + mq^2}{2} - 1\right)^2$. [II. 6]

If then $mp^2 \cdot mq^2 + \left(\frac{mp^2 - mq^2}{2} - 1\right)^2$ is also equal to $\left(\frac{mp^2 + mq^2}{2} - 1\right)^2$, we must have

$$(mp^2 - 2) mq^2 = mp^2 \cdot mq^2,$$

or

$$mp^2 - 2 = mp^2 :$$

which is impossible.

  1. If $mp^2 \cdot mq^2 + \left(\frac{mp^2 - mq^2}{2} - 1\right)^2 < \left(\frac{mp^2 + mq^2}{2} - 1\right)^2$, suppose it equal to $\left(\frac{mp^2 + mq^2}{2} - r\right)^2$.

But [II. 6] $(mp^2 - 2r) mq^2 + \left(\frac{mp^2 - mq^2}{2} - r\right)^2 = \left(\frac{mp^2 + mq^2}{2} - r\right)^2$.

Therefore

$$(mp^2 - 2r) mq^2 + \left(\frac{mp^2 - mq^2}{2} - r\right)^2 = mp^2 \cdot mq^2 + \left(\frac{mp^2 - mq^2}{2} - 1\right)^2 :$$

which is impossible.

Hence all three hypotheses are false, and the sum of the squares $mp^2 \cdot mq^2$ and $\left(\frac{mp^2 - mq^2}{2} - 1\right)^2$ is not square.