Substitute (y2+2py)/a for x in the second and third expressions. must both be squares; or, if we multiply the first expression by → and the second by g' (so as to make the absolute terms the same), we have to solve the double-equation This has to be separated into two factors of the form λy, μy + v, where > must be equal to 297 (in order that, when } {(λ + μ) y + v} is squared and equated to the first, or when {(λ − μ) y − v} is equated to the second, of the two expressions, the absolute terms q may cancel each other). A different separation into factors is possible if b/a and c/a are both squares; but otherwise, as Fermat says, the method gives only one solution in the first instance; the above difference must necessarily be split into the factors gr p(br2-cq2)y and y+2 agr ·y+2qr. — cq2) agr + = {3' (aq2r3 + br2p3 — cp°q3) + gr. apqr Squaring this and equating it to (y2+ 2py) + gr, we have a (aq°r* + br°p2 - cp°q°) + qr}" = br* y* + apqr a br} a aqir2+brip2 - cpq" -y • {2pt; " _ aq2r2 + br2p2. a 4a3p3qï‚a {(aq1r" + br2p3 — cp3q")" — 4abp3q"r"} ap (br°p° + cp°q3 · • aq3r") that is, y (a°q^r^ + b2r^p^ + ¿pʻq^ — zbcpʻq2r2 – 2capʻqʻr2 – zabpʻqr) or y = = 4apq3r3 (br2p2 + cp3q3 — aq3r), 4apq3r2 (br2p2 + cp3q3 — aq3r3) a2q*r+b2r*p* + ¿pʻq — 2bcpʻq°r2 – 2cap3qʻr2 – zabp3q3r whence * (= y2+apy) is found. Exx. from the Inventum Novum. Here a = 1, b = 3, c = 2, p = 1, 9 = 2, r=3; therefore y = = = − 36. 46 + 9 . 81 + 4. 16 ̄ ̄863' and x = y + 2y = (163)2 + 2 (}; }) = 202788 The disadvantage of the method is that it leads so soon to such very large numbers. Other examples from the Inventum Novum are the following, which, like those above given, can be readily solved ab initio without using the above general formula. Put x=y+ay, and substituting in the second and third expressions we have only to solve the double-equation = The difference = 3(y+2y) 3y (y + z). Equate the square of half the difference of the factors to the smaller expression; thus ( y − 1)2 = 2y2 + 4y + 1, In this case we put x=y2+6y, and we have to solve 3(y+6y)+9=v2 5(+6y)+9= w3] The difference = 2 ( y3 + 6y) = 2y (y + 6); we then have 48 If we assume x=ya + 2y, we find y = and x = 11 There are two other problems of the same sort which are curiously enunciated. (6) "To find three cubes such that, if we add their sum to numbers proportional to the cubes respectively, we may have three squares." What Fermat really does is to take three cubes (a3, ¿3, c3) such that their sum is a square (this is necessary in order to make the term independent of x in each of the three expressions a square) and then to assume ax, bx, cx for the numbers proportional to the cubes. He takes as the cubes 1, 8, 27, the sum of which is 36. Thus we have the triple-equation 36+ x = u2 We put x=+ 12y in order to make the first expression a square. Then, solving the double-equation (7) "To find three different square numbers such that, if we add to them respectively three numbers in harmonic progression, the three resulting numbers will be squares." Fermat first assumes three square numbers 1, 4, 16 and then takes 2x, 3x, 6x as the required numbers in harmonic progression. (He observes that, of the three numbers in harmonic progression, the greatest must be greater than the sum of the other two.) We thus have the triple-equation I + 2x = 2a 4+3x= v3 16+6x= wa or, if we make the absolute terms the same square, 16+ 32x = u'a 16+12x = v 16+ 6x = w2 Making the last expression a square by putting y+y for x, we solve as usual and obtain y=-913 and x = } (y2 + 8y) = 12000 Fermat observes that triple-equations of the form x2 + x = u3 x2 + 5x = w3 656 can be similarly solved, because they can be reduced to the above linear form by putting x = 1/y and multiplying up by y We put y = 23+ 28 and, solving the double-equation (3) "To find three square numbers such that, if we add their sum to each of their roots respectively, we obtain a square." Choose, says Fermat, three squares such that their sum is a square and such that the root of the greatest is greater than the sum of the roots of the other two (the reason for this last condition will shortly appear); e.g. let the squares be 4, 36, 81, the sum of which is 121. Let 4x3, 36x3, 81x be the three square numbers required; therefore 121.x2 + 2x = u3 Fermat actually used his triple-equations for the purpose, mainly, of extending problems in Diophantus where three numbers are found satisfying certain conditions so as to find four numbers satisfying like conditions. The cases which occur are in his notes to the problems III. 15, IV. 19, 20, V. 3, 27, 28; they are referred to in my notes on those problems. De Billy observes (what he says Fermat admitted he had not noticed) that the method fails when, the absolute terms being the same square, the coefficient of x in one of the linear expressions to be made squares is equal to the sum of the coefficients of x in the other two. Thus suppose that I + 2x, I + 3x, 1+5x have to be made squares. To make the first expression a square put x=2y+2y. The other expressions then become or 1 + 6y+ 6y2, 1 + 10y + 10y2. The difference is 4y2 + 4y = 2y (2y + 2), and the usual method gives so that y = (2y + 1)2 = 10y3 + 10y + 1, 6y2+6y= 0, - 1, and consequently x = 2y2 + 2y = 0. It does not however follow, says De Billy, that a set of expressions so related cannot be made squares by one value of x. Thus 15x, 1 + 16x and 1+21x are all squares if x = 3. the squares being 16, 49, 64. He adds (§ 11) that we must observe with Fermat" that the triple-equation I + x = u2 I + 2x = v2 1 +3x=w3 not only cannot be solved by the above method, but cannot be solved at all, because "there cannot be four squares in arithmetical progression," which however would be the case if the above equations had a solution and we took 1 for the first of the four squares. I |