SUPPLEMENT

ADDITIONAL NOTES, THEOREMS AND PROBLEMS BY FERMAT, TO WHICH ARE ADDED SOME SOLUTIONS BY EULER

I HAVE generally referred to the notes of Fermat, and allied propositions of his, on the particular problems of Diophantus which were the occasion of such notes, illustrations or extensions; but there are some cases where the notes would have been of disproportionate length to give in the places where they occur. Again, some further explanations and additional theorems and problems given by Fermat are not in the notes to Diophantus but elsewhere, namely in his correspondence or in the Doctrinae Analyticae Inventum Novum of Jacques de Billy "based on various letters sent to him from time to time by Pierre de Fermat" and originally included at the beginning of the 2nd (1670) edition of Bachet's Diophantus (the Inventum Novum is also published, in a free French translation by Tannery, in Oeuvres de Fermat, Vol. 111. pp. 323-398). Some of these theorems and problems are not so closely connected with particular problems in Diophantus as to suggest that they should be given as notes in one place rather than another. In these circumstances it seemed best to collect the additional matter at the end of the book by way of Supplement.

In the chapter on the Porisms and other assumptions in Diophantus (pp. 106-110 above) I quoted some famous propositions of Fermat on the subject of numbers which are the sums of two, three or four square numbers respectively. The first section of this Supplement shall be devoted to completing, so far as possible, the story of Fermat's connexion with these theorems.

SECTION I.

ON NUMBERS SEPARABLE INTO INTEGRAL SQUARES.

As already noted, Fermat enunciated, on Diophantus IV. 29, a very general theorem of which one part states that Every number is either a square or the sum of two, three or four squares. We shall return to this later, and shall begin with the case of numbers which are the sum of two squares.

I.

On numbers which are the sum of two squares.

I may repeat the beginning of the note on III. 19 already quoted (p. 106). "A prime number of the form 4n+ 1 is the hypotenuse of a right-angled triangle in one way only, its square is so in two ways, its cube in three, its biquadrate in four ways, and so on ad infinitum.

"The same prime number 42 + 1 and its square are the sum of two squares in one way only, its cube and its biquadrate in two ways, its fifth and sixth powers in three ways, and so on ad infinitum.

"If a prime number which is the sum of two squares be multiplied into another prime number which is also the sum of two squares, the product will be the sum of two squares in two ways; if the first prime be multiplied into the square of the second prime, the product will be the sum of two squares in three ways; if the first prime be multiplied into the cube of the second, the product will be the sum of two squares in four ways, and so on ad infinitum."

Before proceeding further with this remarkable note, it is natural to ask how Fermat could possibly have proved the general proposition that (a) Every prime number of the form 4n+1 is the sum of two square numbers, which was actually proved by Euler'. Fortunately we have in this case a clear statement by Fermat himself of the line which his argument took. In his "Relation des nouvelles découvertes en la science des nombres" sent by Fermat to Carcavi and shortly after (14 August, 1659) communicated by the latter to Huygens, Fermat begins by a reference to his method of proof by indefinite diminution (descente infinie or indéfinie) and proceeds thus: "I was a long time before I was able to apply my method to affirmative questions because the way and manner of getting at them is much more difficult than that which I employ with negative theorems. So much so that, when I had to prove that every prime number of the form 4n+ 1 is made up of two squares, I found myself in a pretty fix. But at last a certain reflection many times repeated gave me the necessary light, and affirmative questions yielded to my method, with the aid of some new principles by which sheer necessity compelled me to supplement it. This development of my argument in the case of these affirmative questions takes the following line: if a prime number of the form 4n+selected at random is not made up of two squares, there will exist another prime number of the same sort but less than the given number, and again a third still smaller and so on, descending ad infinitum, until you arrive at the number 5 which is the smallest of all numbers of

1 Novi Commentarii Academiae Petropolitanae 1752 and 1753, Vol. IV. (1758), pp. 3-40, 1754 and 1755, Vol. v. (1760), pp. 3-58- Commentationes arithmeticae collectae, 1849, 1. pp. 155-173 and pp. 210-233.

2 Oeuvres de Fermat, II. p. 432.

the kind in question and which the argument would require not to be made up of two squares, although, in fact, it is so made up. From which we are obliged to infer, by reductio ad absurdum, that all numbers of the kind in question are in consequence made up of two squares."

The rest of the note to Diophantus III. 19 is as follows.

"From this consideration it is easy to deduce a solution of the problem "To find in how many ways a given number can be the hypotenuse of a right-angled triangle.

"Take all the primes of the form 4n+ 1, e.g. 5, 13, 17, which measure the given number.

"If powers of these primes measure the given number, set out the exponents of the powers; e.g. let the given number be measured by the cube of 5, the square of 13, and by 17 itself but no other power of 17; and set out the exponents in order, as 3, 2, 1.

"Take now the product of the first of these and twice the second, and add to the product the sum of the first and second: this gives 17. Multiply this by twice the third exponent and add to the product the sum of 17 and the third exponent: this gives 52, which is the number of the different rightangled triangles which have the given number for hypotenuse. [If a, b, c be the exponents, the number of the triangles is 4abc + 2 (bc + ca + ab)+a+b+c.] We proceed similarly whatever the number of divisors and exponents.

"Other prime factors which are not of the form 4n+1, and their powers, do not increase or diminish the number of the right-angled triangles which have the given hypotenuse.

"PROBLEM I. To find a number which is a hypotenuse in any assigned number of ways.

"Let the given number of times be 7. We double 7: this gives 14. Add 1, which makes 15. Then seek all the prime numbers which measure it, e. 3 and 5. Next subtract 1 from each and bisect the remainders. This gives 1 and 2. [In explanation of the process it is only necessary to observe that, for example, 2 {4abc + 2 (bc + ca + ab) + a + b + c} + 1 is equal to (2a + 1) (2b+1) (2c+ 1), and so on.] Now. choose as many prime as there are numbers in the result just arrived at, ie. in this case two. Give to these primes the exponents 1, 2 respectively and multiply the results, ie. take one of the primes and multiply it into the square of the other.

numbers of the form 4 +

"It is clear from this that it is easy to find the smallest number which is the hypotenuse of a right-angled triangle in a given number of ways." [Fermat illustrates the above further in a letter of 25 December 1640

to Mersenne'.

To find a number which is the hypotenuse of 367 different right-angled triangles and no more.

1 Oeuvres de Fermat, 11. pp. 214 sq.

Double the number and add 1; this gives 735. Take all the divisors which are prime numbers: these are 3, 5, 7, 7. Subtract 1 from each and then divide by 2; this gives 1, 2, 3, 3. We have then to take four prime numbers of the form 42 + 1 and give them 1, 2, 3, 3 respectively as exponents. The product of these powers is the number required.

To find the least such number, we must take the four least primes of the form 4n+ 1, i.e. 5, 13, 17, 29, and we must give the smallest of them, in order, the largest exponent; i.e. we must take 53, 133, 172 and 29 in this case, and the product of these four numbers is the least number which is the hypotenuse of 367 right-angled triangles and no more.

If the double of the given number + 1 is a prime number, then there is only one possible divisor. Suppose the given number is 20; the double of it plus 1 is 41. Subtracting unity and bisecting, we have 20, so that the number to be taken is some prime number of the form 4 + 1 to the power of 20.]

"PROBLEM 2. To find a number which shall be the sum of two squares in any assigned number of ways.

"Let the given number, be 10. Its double is 20, which, when separated into its prime factors, is 2.2.5. Subtract 1 from each, leaving 1, 1, 4. Take three different prime numbers of the form 4n+ 1, say 5, 13, 17, and multiply the biquadrate of one (the exponent being 4) by the product of the other two. The result is the required number.

"By means of this it is easy to find the smallest number which is the sum of two squares in a given number of ways.

"In order to solve the converse problem,

"To find in how many ways a given number is the sum of two squares, "let the given number be 325. The prime factors of the form 4n+ 1 contained in this number are 5, 13, the latter being so contained once only, the former to the second power. Set out the exponents 2, 1. Multiply them and add to the product the sum of the two: this gives 5. making 6, and take the half of this, namely 3. This is the number of ways in which 325 is the sum of two squares.

Add 1,

"If there were three exponents, as 2, 2, 1, we should proceed thus. Take the product of the first two and add it to their sum: this gives 8. Multiply 8 into the third and add the product to the sum of 8 and the third: this gives 17. Add 1, making 18, and take half of this or 9. This is the number of ways in which the number taken in this second case is the sum of two squares. [If a, b, c be the three exponents, the number of ways is {abe+ (bc + ca + ab) + (a+b+c) + 1} provided that the number represented by this expression is an integer.]

"If the last number which has to be bisected should be odd, we must subtract 1 and take half the remainder.

"But suppose we are next given the foilowing problem to solve: "To find a whole number which, when a given number is added to it, becomes a square, and which is the hypotenuse of any assigned number of right-angled triangles.

"This is difficult. Suppose eg. that a number has to be found which is a hypotenuse in two ways and which, when 2 is added to it, becomes

a square.

"The required number will be 2023, and there are an infinite number of others with the same property, as 3362 etc."

2. On numbers which cannot be the sum of two squares.

In his note on Diophantus v. 9 Fermat took up a remark of Bachet's to the effect that he believes it to be impossible to divide 21 into two squares because "it is neither a square nor by its nature made up of two squares." Fermat's note was: "The number 21 cannot be divided into two squares (even) in fractions. That I can easily prove. And generally a number divisible by 3 which is not also divisible by 9 cannot be divided into two squares either integral or fractional."

He discusses the matter more generally in a letter of August 1640 to Roberval'.

"I have made a discovery à propos of the 12th [9th] proposition of the fifth Book of Diophantus (that on which I have supplied what Bachet confesses that he did not know and at the same time restored the corrupted text, a story too long to develop here). I need only enunciate to you my theorem, while reminding you that I proved some time ago that

"A number of the form 4n-1 is neither a square nor the sum of two squares, either in integers or fractions."

[This proposition was sent by Mersenne to Descartes, on 22 March 1638, as having been proved by Fermat.]

"For the time I rested there, although there are many numbers of the form 47 + 1 which are not squares or the sums of squares either, e.g. 21, 33, 77, etc., a fact which made Bachet say on the proposed division of 21 into two squares 'It is, I believe, impossible since 21 is neither a square nor by its nature made up of two squares,' where the word reor (I think) clearly shows that he was not aware of the proof of the impossibility. This I have at last discovered and comprehended in the following general proposition.

“If a given number is divided by the greatest square which measures it, and the quotient is measured by a prime number of the form 4n − 1, the given number is neither a square nor the sum of two squares either integral or fractional.

1 Oeuvres de Fermat, 11. pp. 203-4.