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A291591
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Numbers k such that there exist exactly five distinct Pythagorean triangles, at least one of them primitive, with area k.
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1
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OFFSET
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1,1
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COMMENTS
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I solve x^2 + 3*y^2 = (2*r)^2 over the positive integers. q, r, q-p and p are the y-coordinates in the first quadrant. Area = q*r*(q-p)*p. There are three Pythagorean triangles with this area. j, x, y with x > y and Area = j^2*x*y*(x-y)*(x+y) gives the area of an Pythagorean triangle.
Example: r = 169 in x^2 + 3*y^2 = (2*169)^2 gives q = 176, r = 169, q-p = 161 and p = 15;
k = q*r*(q-p)*p = 176*169*161*15 = 71831760.
j = 26, x = 23, y = 12 and j = 26, x = 28, y = 5 gives two Pythagorean triangles with k = 71831760;
k = 676*23*12*11*35 = 71831760 and k = 676*28*5*23*33 = 71831760.
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LINKS
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EXAMPLE
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p^2 - p*q + q^2 = r^2;
p = 115, q = 448, q-p = 333, r = 403;
k = p*q*(q-p)*r = 115*448*333*403 = 6913932480.
x = 414, y = 104 and x = 558, y = 40 gives the same area.
k = x*y*(x-y)*(x+y) = 414*104*310*518 = 6913932480.
k = x*y*(x-y)*(x+y) = 558*40*518*598 = 6913932480.
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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