Polynomial's family!

Algebra Level 4

Let f ( x ) = a x 4 + b x 2 + 3 x + 7 f\left( x \right) = a{ x }^{ 4 } + b{ x }^{ 2 } + 3x + 7 and f ( − 4 ) = 2286 f(-4) = 2286 and f ( 4 ) = N f(4) = N . What is the number of ways in which N N can be resolved as a product of two divisors which are relatively prime?

17 18 16 15

Priyanshu Mishra by Priyanshu Mishra , 20, India

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1 solution

Patrick Corn
Jan 8, 2018

First, N − 2286 = f ( 4 ) − f ( − 4 ) = 3 ( 4 − ( − 4 ) ) = 24 , N-2286 = f(4)-f(-4) = 3(4-(-4)) = 24, so N = 2310 = 2 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 11. N = 2310 = 2\cdot 3 \cdot 5 \cdot 7 \cdot 11. This has 32 32 positive divisors, and they split into 16 \fbox{16} pairs of relatively prime divisors which multiply together to give 2310. 2310.

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jUST CANNOT STOP LAUGHING AT your solution. Its unbelievably the best and good looking solution. Thanks a lot .

Priyanshu Mishra - 3 years, 5 months ago

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