Еquidistant polynomials with integer roots

Calculus Level 5

There are pairs of polynomials with n n integer roots such that P n ( x ) = Q n ( x ) + const P_{n}(x)=Q_{n}(x)+\text{const} .

For example (see image), P 7 ( x ) = x 7 + 7 x 6 − 4193 x 5 − 21035 x 4 + 4674544 x 3 + 14107828 x 2 − 1204669152 x − 8197862400 P_{7}(x)=x^7+7x^6-4193x^5-21035x^4+4674544x^3+14107828x^2-1204669152x-8197862400 .

Find Q 7 ( 0 ) Q_{7}(0) .

D. G. Fon-Der-Flaass talked about this problem.

Bonus: How generate this pairs?

My examples for n = 8 n=8

P 8 ( x ) = ( x 2 − 5 0 2 ) ( x 2 − 3 6 2 ) ( x 2 − 3 1 2 ) ( x 2 − 7 2 ) , Q 8 ( x ) = ( x 2 − 4 9 2 ) ( x 2 − 4 1 2 ) ( x 2 − 2 0 2 ) ( x 2 − 1 8 2 ) P 8 ( x ) = ( x 2 − 5 0 2 ) ( x 2 − 4 2 2 ) ( x 2 − 3 2 2 ) ( x 2 − 4 2 ) , Q 8 ( x ) = ( x 2 − 4 8 2 ) ( x 2 − 4 6 2 ) ( x 2 − 1 0 2 ) ( x 2 − 2 8 2 ) P 8 ( x ) = ( x 2 − 4 7 2 ) ( x 2 − 4 3 2 ) ( x 2 − 1 6 2 ) ( x 2 − 1 5 2 ) , Q 8 ( x ) = ( x 2 − 4 8 2 ) ( x 2 − 4 1 2 ) ( x 2 − 2 3 2 ) ( x 2 − 5 2 ) P 8 ( x ) = ( x 2 − 2 5 2 ) ( x 2 − 2 1 2 ) ( x 2 − 1 6 2 ) ( x 2 − 2 2 ) , Q 8 ( x ) = ( x 2 − 2 4 2 ) ( x 2 − 2 3 2 ) ( x 2 − 1 4 2 ) ( x 2 − 5 2 ) P 8 ( x ) = ( x 2 − 3 4 2 ) ( x 2 − 2 5 2 ) ( x 2 − 2 4 2 ) ( x 2 − 7 2 ) , Q 8 ( x ) = ( x 2 − 3 2 2 ) ( x 2 − 3 1 2 ) ( x 2 − 1 5 2 ) ( x 2 − 1 4 2 ) \begin{array} {ll} P_{8}(x) =(x^2-50^2)(x^2-36^2)(x^2-31^2)(x^2-7^2), & Q_{8}(x)=(x^2-49^2)(x^2-41^2)(x^2-20^2)(x^2-18^2) \\ P_{8}(x)=(x^2-50^2)(x^2-42^2)(x^2-32^2)(x^2-4^2), & Q_{8}(x)=(x^2-48^2)(x^2-46^2)(x^2-10^2)(x^2-28^2) \\ P_{8}(x)=(x^2-47^2)(x^2-43^2)(x^2-16^2)(x^2-15^2), & Q_{8}(x)=(x^2-48^2)(x^2-41^2)(x^2-23^2)(x^2-5^2) \\ P_{8}(x)=(x^2-25^2)(x^2-21^2)(x^2-16^2)(x^2-2^2), & Q_{8}(x)=(x^2-24^2)(x^2-23^2)(x^2-14^2)(x^2-5^2) \\ P_{8}(x)=(x^2-34^2)(x^2-25^2)(x^2-24^2)(x^2-7^2), & Q_{8}(x)=(x^2-32^2)(x^2-31^2)(x^2-15^2)(x^2-14^2) \end{array}


The answer is 5769691200.

Yuriy Kazakov by Yuriy Kazakov , 61, Russia

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1 solution

Patrick Corn
Oct 4, 2020

From a computer search (cheating and estimating the roots from the picture!), I get P 7 ( x ) = ( x − 50 ) ( x − 32 ) ( x − 23 ) ( x + 8 ) ( x + 14 ) ( x + 39 ) ( x + 51 ) Q 7 ( x ) = ( x − 49 ) ( x − 37 ) ( x − 12 ) ( x − 6 ) ( x + 25 ) ( x + 34 ) ( x + 52 ) \begin{aligned} P_7(x) &= (x-50)(x-32)(x-23)(x+8)(x+14)(x+39)(x+51) \\ Q_7(x) &= (x-49)(x-37)(x-12)(x-6)(x+25)(x+34)(x+52) \end{aligned} so Q 7 ( 0 ) = 5769691200. Q_7(0) = 5769691200.

This seems like a difficult problem in general--we've got six equations in seven unknowns, with a known integral solution, and we need to find another integral solution.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

How would you have solved this question if the image were not provided?

Pi Han Goh - 8 months, 1 week ago

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