## RE: [PrimeNumbers] Quiz day soon

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• Here is a solution in integers for which b,c,n, x, y, and z are all different. There are an infinite number of solutions, of course. b = -1 c = 1 n = 0 x =
Message 1 of 3 , Dec 4, 2003
Here is a solution in integers for which b,c,n, x, y, and z are all different. There are an infinite number of solutions, of course.

b = -1
c = 1
n = 0
x = -16
y = 2
z = 81

Paul

> -----Original Message-----
> From: paulmillscv [mailto:paulmillscv@...]
> Sent: 04 December 2003 14:25
> Subject: [PrimeNumbers] Quiz day soon
>
> Hello to Primers,
> Xmas Quiz day is Monday December 15th. The quiz link will be
> posted then and available on all time zones on the 16th. The
> questions will be difficult but not without huge clues as to the
> solution. As a warm up can you find a solution (non-trivial) to this
> question.
>
> Find 3 numbers in Arithmetic Progression A=cn+b B=c(n+1) +b and C=c
> (n+2) +b so that xA^3 , yB^3 and zC^3 are all FOURTH powers. I.e.
> xA^3 = D^4 yB^3 = E^4 and zC^3 = F^4. All letters have integer
> values.
>
> That should keep you occupied until 15th December!
>
> Regards,
>
> Paul Mills
> Kenilworth,
>
>
>
>
>
>
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• I m still trying to figure out what consitutes a non-trivial solution. x=A y=B z=C is a solution for any choices of A, B, C - having nothing to do with their
Message 2 of 3 , Dec 5, 2003
I'm still trying to figure out what consitutes a non-trivial solution.

x=A
y=B
z=C

is a solution for any choices of A, B, C - having nothing to do with
their arithmetic progression. That's surely trivial.

Perhaps the x, y, and z are supposed to be given as fixed values?
That would the problem more difficult.

wrote:
> Hello to Primers,
> Xmas Quiz day is Monday December 15th. The quiz link will be
> posted then and available on all time zones on the 16th. The
> questions will be difficult but not without huge clues as to the
> solution. As a warm up can you find a solution (non-trivial) to this
> question.
>
> Find 3 numbers in Arithmetic Progression A=cn+b B=c(n+1) +b and C=c
> (n+2) +b so that xA^3 , yB^3 and zC^3 are all FOURTH powers. I.e.
> xA^3 = D^4 yB^3 = E^4 and zC^3 = F^4. All letters have integer
> values.
>
> That should keep you occupied until 15th December!
>
> Regards,
>
> Paul Mills
> Kenilworth,
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