 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  All Primes (Posted on 2018-11-17) f(x) = x5 + 5x4 + 5x3 + 5x2 + 1
g(x) = x5 + 5x4 + 3x3 - 5x2 -1

For the above two functions find the sum of all prime numbers p for which there exists a natural number 0 <= x < p such that both f(x) and g(x) are divisible by p.

 No Solution Yet Submitted by Danish Ahmed Khan No Rating Comments: ( Back to comment list | You must be logged in to post comments.) solution | Comment 1 of 2
p will divide the difference f(x) - g(x) = 2x^3 + 10x^2 + 2.

Both functions are always odd for integers so they're divisible only by odd primes and the factor 2 is irrelevant.  Divide it out and subtract the result from f(x) to get x^3(x+4)(x+1).

x<p so the factor x^3 is irrelevant too.  Then the possibilities are  x=-4 or x=-1 mod p.

Apply those values to find f(x)=5 or 17 mod p and g(x)=-5 or -17 mod p.  As the residues are prime, only p=5 or p=17 work.

Checking, f(4)=2705=5*541, g(4)=2415=5*483,  f(13)=525929=17*30937 and g(13)=519843=17*30579.

The primes are thus 5 and 17 with sum 22.

 Posted by xdog on 2018-11-17 21:05:43 Please log in:

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