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Topic: Composite Numbers (Read 1006 times) |
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NickH
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Composite Numbers
« on: Jan 13th, 2003, 5:59pm » |
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Take any positive composite integer, n.
We have n = ab = cd, where ab and cd are distinct factorizations, and a, b, c, d >= 1.
Show that a² + b² + c² + d² is composite.
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towr
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Re: Composite Numbers
« Reply #1 on: Jan 14th, 2003, 1:18am » |
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on Jan 13th, 2003, 5:59pm, NickH wrote:Take any positive composite integer, n.
We have n = ab = cd, where ab and cd are distinct factorizations, and a, b, c, d >= 1.
Show that a² + b² + c² + d² is composite. |
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if n is odd it's easy: a, b, c and d are odd, and thus a² + b² + c² + d² is even, and thus composite..
for the rest I can't readily think of anything but going to the more general case, and trying all combinations:
n =
(w) * (x*y*z) =
(w*x) * (y*z) =
(w*y) * (x*z) =
(w*z) * (x*y) =
(w*x*y) * (z) =
(w*x*z) * (y) =
(w*y*z) * (x)
(w)^2 + (x*y*z)^2 + (w*x)^2 + (y*z)^2 = (w^2 + y^2 * z^2) * (x^2 + 1)
(w)^2 + (x*y*z)^2 + (w*y)^2 + (x*z)^2 = (w^2 + x^2 * z^2) * (y^2 + 1)
(w)^2 + (x*y*z)^2 + (w*z)^2 + (x*y)^2 = (w^2 + x^2 * y^2) * (z^2 + 1)
(w)^2 + (x*y*z)^2 + (w*x*y)^2 + (z)^2 = (w^2 + z^2) * (x^2 * y^2 + 1)
(w)^2 + (x*y*z)^2 + (w*x*z)^2 + (y)^2 = (w^2 + y^2) * (x^2 * z^2 + 1)
(w)^2 + (x*y*z)^2 + (w*y*z)^2 + (x)^2 = (w^2 + x^2) * (y^2 * z^2 + 1)
(w*x)^2 + (y*z)^2 + (w*y)^2 + (x*z)^2 = (w^2 + z^2) * (x^2 + y^2)
(w*x)^2 + (y*z)^2 + (w*z)^2 + (x*y)^2 = (w^2 + y^2) * (x^2 + z^2)
(w*x)^2 + (y*z)^2 + (w*x*y)^2 + (z)^2 = (y^2 + 1) * (w^2 * x^2 + z^2)
(w*x)^2 + (y*z)^2 + (w*x*z)^2 + (y)^2 = (z^2 + 1) * (w^2 * x^2 + y^2)
(w*x)^2 + (y*z)^2 + (w*y*z)^2 + (x)^2 = (w^2 + 1) * (x^2 + y^2 * z^2)
(w*y)^2 + (x*z)^2 + (w*z)^2 + (x*y)^2 = (w^2 + x^2) * (y^2 + z^2)
(w*y)^2 + (x*z)^2 + (w*x*y)^2 + (z)^2 = (x^2 + 1) * (w^2 * y^2 + z^2)
(w*y)^2 + (x*z)^2 + (w*x*z)^2 + (y)^2 = (w^2 + 1) * (x^2 * z^2 + y^2)
(w*y)^2 + (x*z)^2 + (w*y*z)^2 + (x)^2 = (z^2 + 1) * (w^2 * y^2 + x^2)
(w*z)^2 + (x*y)^2 + (w*x*y)^2 + (z)^2 = (w^2 + 1) * (x^2 * y^2 + z^2)
(w*z)^2 + (x*y)^2 + (w*x*z)^2 + (y)^2 = (x^2 + 1) * (w^2 * z^2 + y^2)
(w*z)^2 + (x*y)^2 + (w*y*z)^2 + (x)^2 = (y^2 + 1) * (w^2 * z^2 + x^2)
(w*x*y)^2 + (z)^2 + (w*x*z)^2 + (y)^2 = (y^2 + z^2) * (w^2 * x^2 + 1)
(w*x*y)^2 + (z)^2 + (w*y*z)^2 + (x)^2 = (x^2 + z^2) * (w^2 * y^2 + 1)
(w*x*z)^2 + (y)^2 + (w*y*z)^2 + (x)^2 = (x^2 + y^2) * (w^2 * z^2 + 1)
which is hardly an elegant solution..
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SWF
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Re: Composite Numbers
« Reply #2 on: Jan 15th, 2003, 6:27pm » |
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If a, b, c, and d share a common factor then the sum of squares is obviously a multiple of this factor.
If they do not share a common factor, let a=A1A2, b=B1B2, c=A1B2, d=A2B1. This satisfies the condition that ab=cd. It is seen that a/c=A2/B2. The numerator and denominator of this fraction are equal to A2 and B2 respectively since they are relatively prime (if they weren't a, b, c, and d would share a common factor). A1=a/A2 and B1=b/B2. The sum of squares can be factored as:
a2+b2+c2+d2=A12A22+B12B22+A12B22+A22B12=(A12+B12)(A22+B2 2)
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