For, otherwise, N must be the product of two integers, one of which has the form 4n' + 1, and the other the form 4n" +3; consequently, since the product of these forms is of the form 4n" +3, it follows that N has the form 4n"" + 3, contrary to the supposition. (6.) If N is of the form 4n + 3, then if it has factors, it may be shown, in a similar way, that it may be resolved into two factors, one of the form 4n" +1, and the other of the form 4n""'+ 3. (7.) If N is of the form 4n + 1, and has factors, then it is clear that a +x and a x may represent two factors into which it can be resolved, such that they must both be of the 4n + 1 or 4n + 3; consequently, by addition, we shall have a + x + a x=2a of the form 4n+ 2 or 4n + 6, and of course a must be an odd number. Hence, we shall have (a + x). (α — x) = a2x2= N, or a2= a2- N, (1), in which a must be an odd number; and since a2= N +2, we have a = √N + x2; consequently, if N is not an exact square, it is clear that a must be greater than the greatest integer in WN. And since a and N are odd numbers, it follows from (1) that x must be an even number. (a In like manner, if N is of the form 4n + 3, we shall have x2 = a2 — N, (2), in which a must be an even number greater than VN (if N is not an exact square), and x must be an odd number. (8.) Because the squares of whole numbers terminate with an even number of naughts, 25, 1, 4, 6, or 9, it follows that those whole numbers which terminate with an odd number of naughts, 2, 3, 7, 8, and 5, without terminating with 25, can not be exact squares. (9.) Supposing n to stand for 0 or a positive integer, then it is plain that any whole number will be of one of the forms 3n, 3n+1, 3n+2, or (which is the same) of one of the forms 3n, 3n1; consequently (by squaring) any integral square must be of one of the forms 9n2, 9n2 ± 6n + 1. Hence, it is clear that if we divide any integral square by 3, the remainder must be 0 or 1; consequently, any integer which, divided by 3, gives 2 for the remainder, can not be an exact square. It may be shown, in like manner, that any integral square, when divided by 4, must give 0 or 1 for the remainder; consequently, any integer whose division by gives 2 or 3 for the remainder, can not be an exact square. It is clear that we may in the same way find the forms of the squares which correspond to the divisors 5, 6, 7, etc. (10.) To perceive the use of what has been done, take the following EXAMPLES. 1. To resolve 15229, which is of the form 4n + 1, into its factors. By extracting the square root of the number, we get 123 for the root of the greatest integral square contained in it; consequently, since the proposed number is of the form 4n +1, we must use an odd number for a in (1), which is greater than 123, so that we assume a = 125 + 2n, on the supposition that n is either 0 or a positive integer. Hence, since N = 15229, (1) gives a2 = (125 + 2n)2 – 15229396 + 500n + 4n2 = 4(99+ 125n+ n), which, by putting 1 for n, gives a2=900=30, or x=30; consequently, from a = 125 + 2n = 127 and z = 30, we get a +x= = 157 and ax=97, which give 15229 157 × 97, and the number has been resolved, as required, since 157 and 97 are prime numbers. 2. To resolve 16891, which is of the form 4n+3, into its factors. Because the root of the greatest square contained in the number is 129, and that the number is of the form 4n + 3 (since a in (2) must be an even number, and greater than 129), we put a = 130+2n; and since N = 16891, (2) gives x2 = (130 + 2n)2 - 168919+520n + 4n', which, by putting n = 0, gives a 9, or 3. x = = Hence, we have a = 130 and x = 3, which give a + x = 133 and a-x=127 for factors of the proposed number. Hence, since 13319 x 7, and that 127, 19, and 7 are prime numbers, it follows that 16891 required. 127 × 19 × 7, as 3. To resolve 11051, which is of the form 4n + 3, into its factors. Since 105 is the root of the greatest square contained in the number, and that the number is of the form 4n+ 3, we pnt a = 106+ 2n, and thence, from (2), get a2= (106 + 2n)2 - 11051 185 + 424n + 4n2. Since 185, divided by 3, gives 2 for the remainder, and that 424 and 4, divided by 3, each give 1 for the remainder, it is easy to see that n must be of the form 3m + 1, since it can not be of either of the forms 3m, 3m + 2. Hence, by putting 3m + 1 for n in the preceding equation, we get a2 = 613 + 1296m + 36m2, in which m must be a positive integer. Putting 1, 2, 3, 4, etc., successively for m, we find that m = 7 gives a2 = 1072, which gives x=107. Because a = 106 +2n = 106+ (3m + 1) × 2 = 150, and x = 107, we have a + x = 257 and a-x=43; consequently, we have 11051 = 257 × 43, as required. Remarks.-1st. It will be perceived that the reason why this is more difficult than the preceding questions, is that the factors are much farther from a ratio of equality. 2d. We have extracted the question from p. 207 of Barlow's Theory of Numbers. Barlow has shown that the number is not a prime, but has not given its factors. 4. To resolve 1082401, which is of the form 4n + 1, into its factors. If we multiply the number by 3, it becomes 3247203, of the form 4n+3; consequently, if we can resolve this number into two factors which are different from the proposed number and 3, we shall of course resolve the given number into factors, as required. Because 3247203 is of the form 4n+3, and that 1801 is the root of the greatest square contained in it, we put a = 1802+2n, and thence, from (2), we get a2 = 1 + 7208n + 4n'; consequently, by putting n = 0, we get x = 1. Hence, because a 1802 and x 1, we have a + x = 1803601 × 3 and a 1801; consequently, we have 1082401 = 1801 × 601, as required. Remarks.-1st. If we multiply 11051 by 5, and proceed in like manner, we immediately get 11051 257 × 43, as in the preceding example. 2d. The reason of the preceding process clearly consists in bringing the sought factors to as near a ratio of equality as possible. 3d. We have taken the example from p. 286 of the Theorie Des Nombres, of Legendre, who has found the factors by a much more operose process. = 5. To resolve 997331, of the form 4n+ 3, into its factors. Since 997331 9982 + 1327, and is of the form 4n + 3, we put a = 1000+2n, and thence (from (2)), get a2 = (1000 +2n) 9973312669+ 4000n + 4n3. Because 2669, 4000, and 4, when divided by 3, severally give 2, 1, and 1 for the remainders, it is easy to perceive that n must be of the form 3m + 1; consequently, putting n = 3m + 1, we shall get a2 = 6673 + 12024m + 36m3. Also, because 6673, 12024, and 36, when divided by 5, severally give 3, 4, and 1 for the remainders, it is easy to perceive that m must be of one of the forms 5m' + 2, 5m' + 3, 5m' + 4. Hence, putting 0, 1, 2, 3, 4, 5, etc., successively, for m' in the preceding forms, we soon find that m' 99 gives m= 5m+3=498, which, being put for m, gives a2 = 6673 + 12024m + 36m2 = 14922769 = 3863, or x = 3863. Hence, a + x = 1000 + 2n + x = 1000 + 2(3m + 1) + x = 3990 + 38637853, and a-x=3990-3863127; consequently, we shall have 997331 7853 x 127, as required. 6. To resolve 5428681, of the form 4n + 1, into its factors. Multiplying the number by 41, it becomes 222575921 = 14918229197; consequently, from (1) we get (14919 +2n)-222575921 = 640 + 59676n + 4n2=4(160+14919n +n). Because 160 + 14919n + n3 must be an exact square, and that 160, 14919, and 1, divided by 3, give 1, 0, and 1 for the remainders, it follows that n must be of the form 3m; consequently, putting 3m for n, we get a2 = 4(160 +44757m +9m2); and in much the same way, it appears that m must be of one of the forms 5m', 5m' + 1, 5m' + 2. Hence, putting 0, 1, 2, 3, 4, etc., for m', successively, in these forms, we soon get m' = 7, which gives m = 5m′ + 1 36, such that a 4(160 +1611252 + 11664) 25483, or x=2548. Hence, from a = 14919 + 2n = 14919 + 6m = 15135 and x = 2548, we get a + x = 17683 and a-x= 12587 = 307 × 41; consequently, 17683 and 307 are the factors of the proposed number. Remarks.-1st. Multiplying the number given in the preceding example by 61, and that in this by 57, and then proceeding as in Ex. 4, we immediately get their factors. 2d. This and the preceding example have been taken from p. 417 of Gauss' Recherches Arithmetiques. (11.) Supposing the odd integer N to have A and B for its factors, and that A is much less than B, and that N=AB is multiplied by the odd integer m, then we shall clearly have (mA + B)2 _ (mA — B)2 = mAB, (3); consequently, a and x in (1) and (2) may be expressed by mA + B and If we divide B by A, it is easy to perceive that such an integer may be found for m, that the corresponding value of mA - B 2 shall be less than + A and numerically less than Hence, we may suppose the determination of the factors A and B of N to be reduced to that of finding m, such that mA - B shall be a minimum, when the sign is neglected. 2 Supposing 2m' to be an integer less than m, and that m is changed to m± 2m', (3) becomes (mA - B 2 (mA + B (mA2 2 ± m'A)2 ± m'A)*= (m ± 2m')AB, (4), in which the differ ence of the successive values of mA + B 2 +m'A and m'A that result from m' = 1, 2, 3, etc., successively, is nu merically equal to A. |