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Old 2006-10-23, 15:55   #3
roger
 
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Oct 2006

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First of all, sorry for the erroneous data to you who have worked on this.
Here are all the ones I have that are proven:

^2 : n^2-x^2 = (n-x)(n+x); n^2+x^2 = (n-x)(n+x)+x
^3 : n^3-n*x^2 = (n-x)(n)(n+x); n^3+nx^2 = (n-x)(n)(n+x)+2n*x^2
^4 : n^4-5n^2*x^2+4x^4 = (n-2x)(n-x)(n+x)(n+2x);
n^4+5n^2*x^2+4x^4 = (n-2x)(n-x)(n+x)(n+2x)+10n*x^2
^5 : n^5-5n^3*x^2+4n*x^4 = (n-2x)(n-x)(n)(n+x)(n+2x);
n^5+5n^3*x^2+4n*x^4 = (n-2x)(n-x)(n)(n+x)(n+2x)+10n^3*x^2

At the moment, ^6 is not on there because, even though I have double and checked again, whenever I enter a value for the equation I came up with (double checking the work behind it too), I get a negative or zero value, or at least ones that don't fit the approximated values. I will continue trying to fix this one, and will write back quicker than last time.

Thanks to all those who attempted and wrote back, and I hope I haven't taken up too much of your time with my bad equations.

Roger

PS - mfgoode - n^2+1 often/always comes up with primes. Will look back into this also

Last fiddled with by roger on 2006-10-23 at 15:56
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