Harold M. Stark,

We see that x = 1, y = 0 satisfies the Diophantine equation

(5) x^2 -1141y^2 = 1

We might ask, does equation (5) have any solution in positive integers? We see from (5) that

x = sqrt(1141y^2 + 1)

Thus the question is: Is 1141 y + 1 ever a perfect square? This may be checked experimentally. It turns out that the answer is no for all positive y less than 1 million. In view of the previous example, perhaps we should experiment further. The answer is still no for all y less than 1 trillion (1 million million, or 10^12). We go overboard and check all y up to 1 trillion trillion (10^24). Again the answer is, no. No one in his right mind would really believe that there could be a positive y such that

x = sqrt(1141y^2 + 1)

is an integer if there is no such y less than 1 trillion trillion. But there is. In fact there are infinitely many of them, the smallest among them having 26 digits.

In 1999 I attended a lecture by Prof. R. Graham, newly of UCSD, the most recent academic employer of Prof. H. Stark. I sent Prof. Graham the above. He answered with a question about a second large number for a situation where only one had been known for many years. To learn more about what happened, in particular how the internet aided communications leading to a new result, please go to Net. For the result (and some questions) please go to New.