and, since

<b, therefore x=b-1-ao, or some

integer number: whence the equation

ax + by = c

is always possible, when a and b are prime to each other, and c> (ab — a − b).

-

Q. E. D.

Cor. The two foregoing propositions are very useful in judging of the possibility, or impossiblity, of indeterminate equations of this kind; and, cons sequently, also, in proposing them, so that they may be within possible limits.

79

CHAP. IV.

On the possible and impossible Forms of Square Numbers, and their Application to Numerical Propositions.

PROP. I.

42. Every square mumber is of one of the forms 4n, or 4n+ 1.

For every number, being either even or odd, is of one of the forms 2n, or 2n+1; and, consequently, every square number is of one of the forms 4n', or (4n+4n+1); but

4n4n, and

(4no + 4n + 1) = 4(n2 + n) + 1 =4n + 1.

Q. E. D.

Cor. 1. Every square of the form 4n is necessarily even; and every square of the form 4n+1 is evidently odd; therefore, every even square is of the form 4n, and every odd square of the forin

4n+ 1.

appears,

Cor. 2. By the foregoing proposition it that every odd square is of the form 4(n2 + n) + 1; and hence it follows, that it is also of the form 8n+1: for if n be odd, n' is odd, and if n be even, n' is even also; therefore, in both cases, n2+n is even; and, consequently, 4(n + n) +1±8n + 1; that is, every odd square is of the form 8n+1. If,

therefore, a number be of the form 4n+1, but not of the form 8n+1, that number is not a square.

Cor. 3. No numbers of the forms 4n + 2, or 4n+3, can be square numbers. Nor can any numbers of the forms 8n+2, 8n+3, 8n+5, 8n+6, or 8n +7, be square numbers.

Cor 4. The sum of two odd squares cannot form a square, for (4n+ 1) + (4n' + 1)4n+2, which (4n+1)+ cannot be a square (cor. 3).

Cor. 5. An odd square, subtracted from an even square, cannot leave a square remainder. For

4n - (4n' + 1)=4(n − n') − 1 ± 4n+3,

which cannot be a square. Therefore, if the difference of an even and odd square be a square, the odd square must be the greatest.

Cor. 6. If the sum of an even and odd square be a square, the even square must be divisible by 16, or be of the form 4 n'. For all odd squares are of the form 8n+1 (cor. 2); and, therefore, if the even square had only the form 4n", n22 being odd, the sum of the two would be

4n22 + 8n + 1 = 4 (n22 + 2n) + 1;

and since n is odd, (n22 + 2n) is odd also; and, therefore, 4(n2 + 2n) + 1 is not of the form 8n +1; and, consequently, it is not a square (cor. 2).

PROP. II.

43. Every square number is of one of the forms

5n, or 5n1.

For all numbers, compared by modulus 5, are of one of the forms 5n, 5n+1, 5n+ 2; that is, every number is either divisible by 5, or will leave for a

remainder 1, 2, 3, or 4; or, which is the same, ±1, or 2: and, consequently, all square numbers are of one of the following forms:

5n±2, 25n2±20n +4 ± 5n+45n−1.

Consequently, all square numbers are of one of the forms 5n, or 5n ±1. — Q. E. D.

Cor. 1. If a square number be divisible by 5, it is also divisible by 25; and, if a number be divisible by 5, and not by 25, it is not a square.

Cor. 2. No number of the form 5n+2, or 5n+3, is a square number.

Cor. 3. If the sum of two squares be a square, one of the three is divisible by 5, and, consequently, also by 25. For all the possible combinations of the three forms 5n, 5n+1, and 5n-1, are as follows:

(5n + 1 ) + (5n' + 1 ) ± 5n + 2,

(5n − 1) + (5n' −1)5n-25n+3,

Now of these six forms, the latter four have one of the squares divisible by 5, and, therefore, also by 25 (cor. 1). And the two first are each impossible forms for square numbers; that is, neither of these two combinations can produce squares: therefore, if the sum of two squares be a square, one of the three squares is divisible by 25.

Cor. 4. By means of the two foregoing propositions, and their corollaries, it appears, that no number contained under a repetend digit can be a square number.

For every number expressed by a repetend digit is equal to the same number of repetend units, multiplied by the particular digit under the repetend of which the number is contained.

But every repetend of units is of the form

100n+114n+35n+ 1;

and it is only necessary to show, that no number of the form 4n+3, or 5n+ 1, multiplied by any one of the nine digits, can be a square. Now the following products,

(4n+3) × 1,

(4n+3) × 4,

(4n+3) × 9,

cannot produce squares, because one of the factors is a square, and the other not; and, consequently, the product cannot be a square (art. 15).

Again,

(4n+3) × 24n' + 2,

(4n+3) × 5±4n' +3,

(4n+3) × 64n' + 2;

which are all impossible forms for squares (cor. 3, art. 42). And since a repetend unit is likewise of the form 5n+1, we have

(5n + 1) × 3 ± 5n′+3,

(5n+ 1) × 75n' + 2,

(5n+ 1) × 85n'+3,

each of which is an impossible form (cor. 2, art. 43);