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[halloffame]

Tìm các số nguyên dương $x,y,z,t$ đôi một phân biệt thỏa $x^{2}-y^{2}=y^{2}-z^{2}=z^{2}-t^{2}.$

Bài viết đã được chỉnh sửa nội dung bởi halloffame: 12-03-2016 - 16:30

[Visitor]

Tìm các số nguyên dương $x,y,z,t$ đôi một phân biệt thỏa $x^{2}-y^{2}=y^{2}-z^{2}=z^{2}-t^{2}.$

Bài toán kinh điển từ năm $1640$: ko có cấp số cộng nào có bốn số hạng liên tiếp là số chính phương. 

và lời giải hay nhất của nó:

"

No Four Squares In Arithmetic Progression

 

To prove that four consecutive terms in an arithmetic sequence cannot all be squares, suppose there exist four squares A², B², C², D² in increasing arithmetic progression, i.e., we have B²-A² = C²-B² = D²-C². We can assume the squares are mutually co-prime, and the parity of the equation shows that each square must be odd. Hence we have co-prime integers u,v such that A = u-v, C = u+v, u²+v² = B², and the common difference of the progression is (C²-A²)/2 = 2uv.

 

We also have D² - B² = 4uv, which factors as [(D+B)/2][(D-B)/2] = uv. The two factors on the left are co-prime, as are u and v, so there exist four mutually co-prime integers a,b,c,d (exactly one even) such that u = ab, v = cd, D+B = 2ac, and D-B = 2bd. This implies B = ac-bd, so we can substitute into the equation u²+v² = B²  to give (ab)²+(cd)² = (ac-bd)². This quadratic is symmetrical in the four variables, so we can assume c is even and a, b, d are odd. From this quadratic equation we find that c is a rational function of the square root of  a⁴-a²d²+d⁴, which implies there is an odd integer m such that  a⁴-a²d²+d⁴= m².

 

Since a and d are odd there exist co-prime integers x and y such that a² = k(x+y) and d² = k(x-y), where k = ±1. Substituting into the above quartic gives x²+3y² = m², from which it's clear that y must be even and x odd. Changing the sign of x if necessary to make m+x divisible by 3, we have 3(y/2)² = [(m+x)/2][(m-x)/2], which implies that (m+x)/2 is three times a square, and (m-x)/2 is a square. Thus we have co-prime integers r and s (one odd and one even) such that (m+x)/2 = 3r², (m-x)/2 = s², m = 3r²+s², x = 3r²-s², and y = ±2rs.

 

Substituting for x and y back into the expressions for a² and d² (and transposing if necessary) gives a² = k(s+r)(s-3r) and d² = k(s-r)(s+3r). Since the right hand factors are co-prime, it follows that the four quantities (s-3r), (s-r), (s+r), (s+3r) must each have square absolute values, with a common difference of 2r. These quantities must all have the same sign, because otherwise the sum of two odd squares would equal the difference of two odd squares, i.e., 1+1 = 1-1 (mod 4), which is false.

 

Therefore, we must have |3r| < s, so from m = 3r² + s² we have 12r² < m. Also the quartic equation implies m < a² + d², so we have the inequality |2r| < |SQRT(2/3)max(a,d)|. Thus we have four squares in arithmetic progression with the common difference |2r| < |2abcd|, the latter being the common difference of the original four squares. This contradicts the fact that there must be a smallest absolute common difference for four squares in arithmetic progression, so the proof is complete.□ "

Bài viết đã được chỉnh sửa nội dung bởi Visitor: 18-03-2016 - 08:13