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

A

B

even number, or an odd number from an odd number, the remainder is even,

therefore the remainder AC is even.

Let AC be bisected at D.

[IX. 24, 26]

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.

[11.6]

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 1. 47, is as follows.

Take two similar plane numbers, e.g. mnp2, mnq3, which are either both even or both odd, so that their difference is divisible by 2.

Now the product of the two numbers, or m2n2p'q, is square,

and, by II. 6,

2

mnp2 . mnq2 + (mnp2 = mnq3)2 = (mnp2 + mnq2)",

2

=

[Lemmas 1, 2

[IX. 1]

so that the numbers mnpq, (mnp2 - mnq") satisfy the condition that the sum of their squares is also a square number.

It is also clear that (mnp2 + mnq3), 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.

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;

Lemma 2]

LEMMAS TO PROPOSITIONS 29, 30

65

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.

I

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

where, as before, mp2, mq2 are both odd or both even.

66

BOOK X

[Lemma 2, x. 29

The number is clearly less than mp3. mq2 + (mp2 — mq”)", i.e. less than

2

2

2

−1),

'mp2 + mq2 \ 2

2

If then the number is square, its side must be greater than, equal to, or less than (mp2+mq2

the number next less than

mp2 + mg3

2

Hence all three hypotheses are false, and the sum of the squares

mp2. mg2 and

'mp2 – mq3

(mp3

2

To find two rational straight lines commensurable in square only and such that the square on the greater is greater than the square on the less by the square on a straight line commensurable in length with the greater.

For let there be set out any rational straight line AB, and two square numbers CD, DE such that their difference CE is not square;

let there be described on AB the semicircle AFB,

[Lemma 1]

therefore the square on DA nas to the square on AF the ratio which the number DC has to the number CE; therefore the square on BA is commensurable with the square on AF

But the square on AB is rational ;

A

therefore the square on AF is also rational; therefore AF is also rational.

And, since DC has not to CE the ratio which a number has to a square number,

neither has the square on BA to the square on AF th which a square number has to a square number; therefore AB is incommensurable in length with AF. Therefore BA, AF are rational straight lines co surable in square only.

And since, as DC is to CE, so is the square on the square on AF,

therefore, convertendo, as CD is to DE, so is the squ AB to the square on BF. [v. 19, Por., III. 3 But CD has to DE the ratio which a square numb to a square number;

therefore also the square on AB has to the square the ratio which a square number has to a square numbe therefore AB is commensurable in length with BF.

And the square on AB is equal to the squares on A therefore the square on AB is greater than the square by the square on BF commensurable with AB.

Therefore there have been found two rational s lines BA, AF commensurable in square only and suc the square on the greater AB is greater than the squa the less AF by the square on BF commensurable in with AB.

Q. E. I