If and are positive integers such that , show that is divisible by 641.
Hi,
I found the same sum as above.
And as for 641, notice that it is prime. So I guess the result of the summation will have a factor 641 in the numerator.
That's all I can guess about.
I tried to compute
But didn't give it enough time (it was late )...
641 is prime, as Moo mentioned, so it does not divide the denominator of our fraction, as the denominator is just 1.2.3. ... .480. Thus, if 641 divides the numerator of our product it will divide every such numerator (this is not clear as 3 divides the numerator but also the denominator in our sum).
I have no idea how it would come up in this problem, but 641 doesn't seem random to me: it reminds me immediately the proof by Euler that the Fermat number is not prime (thus breaking down Fermat's conjecture). Euler showed that 641 divides . I guess this is not a coincidence, but I may be totally wrong. If it may help anyone...