I have a question too.

literka · 2013-06-14, 14:51

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

R.D. Silverman · 2013-06-14, 16:01

Quote:

Originally Posted by literka

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

Clearly, yes.

f(x) = 641, g(x) = 6700417, h(x) = 2^16 seems to work.......

literka · 2013-06-14, 16:25

Of course, you are right. What I meant it is non-trivial 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.

firejuggler · 2013-06-14, 17:04

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'?

literka · 2013-06-14, 17:23

Quote:

Originally Posted by firejuggler

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'?

I am sure you misunderstood my question. The question is to find polynomials not numbers. And not any polynomials, but polynomials satisfying conditions 1-4.
Frankly saying I cannot find anything in your answer which is relating to my question.

firejuggler · 2013-06-14, 17:25

at least we are in miscellanous math, i'm allowed some leeway.

literka · 2013-06-14, 17:52

Quote:

Originally Posted by firejuggler

at least we are in miscellanous math, i'm allowed some leeway.

Yes, you are in miscellanous. It does not mean you are free to write anything you like. If I was a mod, for sure I would ban you. I saw people banned for something less important. You are a fool.

2013-06-14, 14:51	#1
literka Mar 2010 2⁶×3 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

2013-06-14, 16:25	#3
literka Mar 2010 C0₁₆ Posts	Of course, you are right. What I meant it is non-trivial 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 2013-06-14 at 16:35

2013-06-14, 17:04	#4
firejuggler "Vincent" Apr 2010 Over the rainbow 2×7²×29 Posts	f(4)=13 g(1)=2 h(1)=239 239^2+1=13^42 ; 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 2013-06-14 at 17:14

2013-06-14, 17:25	#6
firejuggler "Vincent" Apr 2010 Over the rainbow 2×7²×29 Posts	at least we are in miscellanous math, i'm allowed some leeway.