|
Question 1. Prove that H is not divisible by (x+y):
1. Area H is (z^2-xy) =(x^2+y^2-xy)
2 (x+y) shares no factors with x or y; by definition, those numbers (and also z) are relatively prime. Assume that some factor, k, of x also divides (x+y): then we would have y = (n-m)k and k also divides y, which is contrary to the definition: the same approach applies to (x-y).
3. (x^2+y^2-xy) = (x-y)^2 + xy. The first term on the RHS shares no common factors with x or y, while the second is divisible by either. So the whole is relatively prime to both x and y.
4. Dividing (x^2+y^2-xy)/(x+y) gives, e.g. (3x^2)/(x+y)-2x+y, where the x^2 term is a fraction unless (x+y) = 3; however there is no such PPT. So (x^2+y^2-xy) and (x+y) are also relatively prime, and Question 1 is solved.
Question 2. Prove that there are no (x,y) such that H is a cube.
Preliminary.
5. Assume a putative nth root of H, h, so that (z^2-xy)=(x^2+y^2-xy)=h^n, or xy=z^2-h^n. We know by definition that neither x nor y divides z or z^2. Assume that x or y divided h^n; equivalently, x or y would divide h, multiplied by some other component,k. Then xy = z^2-(kx)^n, so that x(k^nx^(n-1)+y) = z^2 and x divides z, contrary to the definition. Accordingly, neither x nor y divides h; in like manner, substituting nk1 for x and y, it is also clear that x,y and h are, again, relatively prime.
6. By definition, x and y are of opposite parity, and z is odd, so xy is even and h is odd.
Proof.
7. We know that (x+y)(x^2+y^2-xy) is a sum of cubes, x^3+y^3. On the assumption that there is some h^3 that substitutes for the second term, we write (x+y)*h^3= x^3+y^3; equivalently,(x+y)*h^3-x^3 = y^3, so that y(h^3-y^2) = x(x^2-h^3), where (by reason of e.g. (y/x)(h^3-y^2) = x^2-h^3) x must divide (h^3-y^2) and y must divide (x^2-h^3) for the assumption to hold, giving y(kx) = x(k1y) - but in that case k=k1.
8. Therefore h^3 = kx+y^2 = x^2-ky; subtracting one of the two latter expressions from the other gives x(k-x)+y(k+y) = 0, which obviously can hold only where x,y (and accordingly h) are 0, contrary to the definition. Hence there are no (x,y) for which H is a cube, and Question 2 is solved.
Question 3. Prove that if H is a square, H is divisible by 13.
9. This time, the putative square root of H is h, so that (z^2-xy)=(x^2+y^2-xy)=h^2. The methods required to establish the relative primality of all terms have already been developed above.
Preliminary.
10. Let x be the even term. y is odd, so the difference between x and y must be odd; further, assuming that some odd difference greater than (ABS)1, n, divides either x or y, then h^2 = (kn)^2+(kn+n)^2-kn(kn+n) - and it is clear that n must divide both x and y, as well as h^2. It follows that n too, is relatively prime to both x and y.
11. Since x^2 + y^2 = z^2, both z^2 and z are of the form 4k+1. x^2 is divisible by 4, so that x is divisible by 2 or 4. y is of the form (4k+1) or (4k+3). Assume that x were divisible by 2 only; xy would be worth 2, mod4, and h^2 would be worth 2, mod4, which is not possible for a square. So x is divisible by 4.
12. This gives (4k)^2+(4k+n)^2 = z^2 and (4k)^2+(4k+n)^2-4k(4k+n) = h^2 simplifying to 32 k^2+8kn+n^2 = z^2 and 16k^2+4kn+n^2 = h^2; the difference between these two being 16k^2+4kn, equivalent to xy.
Proof
13. Assume that xy may be written as 2ab, and z as (a+b)^2. Using the expansion of (a+b)^2, we now have x^2+y^2=a^2+2ab+b^2 = z^2, and x^2+y^2-2ab as a^2+b^2 = h^2, with z = (a+b), with a,b,h,z all, as usual, relatively prime, and with h of the form (4k+1) as might have been expected, given the context.
14. But all this assumes the correctness of the first moves, i.e. that xy has the exact same factors as 2ab, and that z = (a+b). From x^2+y^2=z^2 we have the Pythagorean form x =2uv, y=u^2-v^2, giving z^2-xy as (u^2+v^2)-2uv(u-v)(u+v). Thus (u^2+v^2)^2=h^2+2uv(u-v)(u+v) and since as has already been shown, all of those terms are relatively prime, the factors of the putative ab are contained in u,v,(u+v) and (u-v).The smallest set compliant with those requirements is {1,4,5,3} giving 2ab = (2*3*4*5) = 120. Crosschecking with 16k^2+4kn gives the all-positive solutions {k,n} as {1,26} and {2,7}. The former produces 16k^2+4kn+n^2 = 796, which is not a square, the latter 169, which is. Further, 169+120 = 289 which is 17^2. It is then easily confirmed that a=5,b=12,h=13,x=8,y=15,z=17, and xy, ie 8*15, is indeed the same as 2ab, ie 2*5*12.
15. Plainly no smaller number than 13 could qualify as h. What about larger numbers? From 32k^2+8kn+n^2 = z^2 and 16k^2+4kn+n^2 = h^2 the form (z^2-n^2) = 2(h^2-n^2) produces the Pell-type equation 2h^2 = n^2+z^2 (here, 2*169 = 289+49), or 2h^2 = (n+z)^2 - 2nz. Checking small values of h <100 using x^2+y^2 = z^2, z^2-xy = h^2 finds:{13,8,15,17}{26,16,30,34}{39,24,45,51}{52,32,60,68}{65,40,75,85}{78,90,48,102} {91,56,105,119}...{169,104,195,221} with the nth set being produced by multiplying each element by n, so that by recursion there are no more solutions to the equation.
16. It follows that the relation x^2+y^2 = z^2, z^2-xy = h^2, where x,y,z are sides of a PPT holds good iff h=13.
|
