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        riddles >> easy >> Composite Numbers
        
(Message started by: NickH on Jan 13th, 2003, 5:59pm)
Title: Composite Numbers
Post by NickH on Jan 13th, 2003, 5:59pm
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.
Title: Re: Composite Numbers
Post by towr on Jan 14th, 2003, 1:18am

on 01/13/03 at 17:59:02, 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.

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..
Title: Re: Composite Numbers
Post by SWF on Jan 15th, 2003, 6:27pm
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+B22)


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