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Old 2006-10-24, 22:50   #4
roger
 
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Oct 2006

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I checked back on n^6, and discovered I'd been thinking backwards, so confused after working pages of bad data that (obviously) gave no results.

For n^6 though, I have come to another stumbling block. All I have discovered is that for at least x<4 (exclusive), the difference between the right hand side of the equation [(n-x)(n+x) for n^2] and the left [n^2*x^2 for n^2] is divisible by 7.

My method for finding n^a+ (probably slow) is to obtain n as well as the right side of the equation, giving a value, and subtracting that value from the left side to obtain a new value (I'll call this b). a is then some form or other (the right-most bit of the right side), but with n^6+, all I know/can see is that b/y is a whole number (and then can be further broken down into a form of some kind, like the other a's)

Can anyone advise?

Thanks,

Roger
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