20130614, 14:51  #1 
Mar 2010
2^{6}×3 Posts 
I have a question too.
Are there three polynomials f,g,h with integer coefficients and
1. f(2)=641 2. g(2)=6700417 3. h(2)=65536=2^16 and 4. f*g=1+h^2 
20130614, 16:01  #2 
Nov 2003
7460_{10} Posts 

20130614, 16:25  #3 
Mar 2010
C0_{16} Posts 
Of course, you are right. What I meant it is nontrivial solution.
For example: degree of f = 9, degree of g=23. I think, I found the way to find f and g. But there is a lot of computations. But it is my blame. I did not notice this solution. In this case I am afraid that there are still some straightforward solutions. Of course it has something to do with factorization of Fermat numbers. Because instead of finding factors we can try to find polynomials (if they exist) with above properties. It may be easier to find polynomials, because we may factor values of polynomials in point, where they are much smaller. Last fiddled with by literka on 20130614 at 16:35 
20130614, 17:04  #4 
"Vincent"
Apr 2010
Over the rainbow
2×7^{2}×29 Posts 
f(4)=13
g(1)=2 h(1)=239 239^2+1=13^4*2 ; does that count or did I misanderstand again? f(2)=65 g(1)= 17 h(1)=268 432^2+1 = 5^3 * 1493 1068^2+ = 5^6 * 73 are these considered ' trivial'? Last fiddled with by firejuggler on 20130614 at 17:14 
20130614, 17:23  #5  
Mar 2010
2^{6}·3 Posts 
Quote:
I am sure you misunderstood my question. The question is to find polynomials not numbers. And not any polynomials, but polynomials satisfying conditions 14. Frankly saying I cannot find anything in your answer which is relating to my question. 

20130614, 17:25  #6 
"Vincent"
Apr 2010
Over the rainbow
2×7^{2}×29 Posts 
at least we are in miscellanous math, i'm allowed some leeway.

20130614, 17:52  #7 
Mar 2010
C0_{16} Posts 
