Interesting. I've solved several more puzzles in the Community group after this one, but I still can't see your solution, Binbin. It's obviously much smaller than mine is, which is why I'm interested in figuring out how you did this.
@ James
To unlock all solutions of this problem, you must solve a newer problem than this one, not older problems.
@James, my solution is very simple, it is pure number.The challenge is from Fermat. Because the big number, it is hard to test. I want use Java's Big Integer, but it has no sqrt operation.
My program gives a wrong answer for n=1, but passes the tests due to insufficient precision. How big is the smallest solution for n=1?
@Tim, it is about 13 digitals. I want to submit another problem use char array as output and avoid to precision.
@James, @Tim, Because of precision, the problem have no meaning, so I have fix the test suits. The output must be char array. Thank you for your support.
I was using the method to parameterize Pythagorean triples (a,b,c) in terms of s and t. Since we're looking at c^4 instead of c^2 (to make sure that c itself is a perfect square), I can see why the numbers got quite large. My solution for n=2 was for larger numbers than my n=3 solution, although even my n=1 solution had the magnitude of a as 10^16 and b was 10^21. I didn't find anything on the order of 10^13.
@James, I think the following link will help you.
https://www.mathpages.com/home/kmath022/kmath022.htm
Very nice link, Binbin. That is was exactly what I was trying to do, only they didn't include the math error I made in my calculations. :)
well, I seem to have missed the smallest solutions...
wow, very impressive! (I get how to obtain the a=n^8 - 16*n^6 ... solution, but how did you get the long-polynomial "if n<2" solution?)
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