with * a* , * b* any coprime integers, * a* > 1 and −* a* < * b* < * a* . (Since * a* * n* − * b* * n* is always divisible by * a* − * b* , the division is necessary for there to be any chance of finding prime numbers. In fact, this number is the same as the Lucas number * U n * (* a* + * b* , * ab* ) , since * a* and * b* are the roots of the quadratic equation * x* 2 − (* a* + * b* )* x* + * ab* = 0 , and this number equals 1 when * n* = 1 ) We can ask which * n* makes this number prime. It can be shown that such * n* must be primes themselves or equal to 4, and * n* can be 4 if and only if * a* + * b* = 1 and * a* 2 + * b* 2 is prime. (Since * a* 4 − * b* 4 / * a* − * b* = (* a* + * b* )(* a* 2 + * b* 2 ) . Thus, in this case the pair (* a* , * b* ) must be (* x* + 1, −* x* ) and * x* 2 + (* x* + 1) 2 must be prime. That is, * x* must be in A027861 .) It is a conjecture that for any pair (* a* , * b* ) such that for every natural number * r* > 1 , * a* and * b* are not both perfect * r* th powers, and −4* ab* is not a perfect fourth power . there are infinitely many values of * n* such that * a* * n* − * b* * n* / * a* − * b* is prime. (When * a* and * b* are both perfect * r* th powers for an * r* > 1 or when −4* ab* is a perfect fourth power, it can be shown that there are at most two * n* values with this property, since if so, then * a* * n* − * b* * n* / * a* − * b* can be factored algebraically) However, this has not been proved for any single value of (* a* , * b* ) .

The record passed one million digits in 1999, earning a $50,000 prize. [6] In 2008 the record passed ten million digits, earning a $100,000 prize and a Cooperative Computing Award from the Electronic Frontier Foundation . [5] * Time * called it the 29th top invention of 2008. [7] Additional prizes are being offered for the first prime number found with at least one hundred million digits and the first with at least one billion digits. [5] Both the $50,000 and the $100,000 prizes were won by participation in GIMPS.