The Hidden Prime

Algebra Level 2

x 2 − P ( x − 2 ) ( x − 3 ) ( x − 5 ) = A x − 2 + B x − 3 + C x − 5 \dfrac{x^2 - P}{(x - 2)(x - 3)(x - 5)} = \dfrac{A}{x - 2} + \dfrac{B}{x - 3} + \dfrac{C}{x - 5}

The equation above represents a partial fraction decomposition for constants A , B , C A,B,C and P P .

What is the smallest value of the prime number P P such that A , B A,B and C C are all integers ?


The answer is 7.

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Worranat Pakornrat by Worranat Pakornrat , 36, Thailand

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1 solution

Relevant wiki: Partial Fractions - Cover Up Rule

By using the cover-up rule for the partial fractions, we can evaluate the values of the numerators A A , B B , & C C as shown:

A = f ( 2 ) ( 2 − 3 ) ( 2 − 5 ) = 4 − P 3 A = \dfrac{f(2)}{(2 - 3)(2 - 5)} = \dfrac{4 - P}{3}

B = f ( 3 ) ( 3 − 2 ) ( 3 − 5 ) = 9 − P − 2 B = \dfrac{f(3)}{(3 - 2)(3 - 5)} = \dfrac{9 - P}{-2}

C = f ( 5 ) ( 5 − 2 ) ( 5 − 3 ) = 25 − P 6 C = \dfrac{f(5)}{(5 - 2)(5 - 3)} = \dfrac{25 - P}{6}

Now we can plug in the possible values of primes, starting from 2 2 onwards:

P = 2 ⇒ A = 4 − 2 3 = 2 3 P = 2 \Rightarrow A = \dfrac{4 - 2}{3} = \dfrac{2}{3} , which is not an integer.

P = 3 ⇒ A = 4 − 3 3 = 1 3 P = 3 \Rightarrow A = \dfrac{4 - 3}{3} = \dfrac{1}{3} , which is not an integer.

P = 5 ⇒ A = 4 − 5 3 = − 1 3 P = 5 \Rightarrow A = \dfrac{4 - 5}{3} = \dfrac{-1}{3} , which is not an integer.

Now P = 7 ⇒ A = 4 − 7 3 = − 3 3 = − 1 ; B = 9 − 7 − 2 = 2 − 2 = − 1 ; C = 25 − 7 6 = 18 6 = 3 P = 7 \Rightarrow A = \dfrac{4 - 7}{3} = \dfrac{-3}{3} = -1; B = \dfrac{9 - 7}{-2} = \dfrac{2}{-2} = -1; C= \dfrac{25 - 7}{6} = \dfrac{18}{6} = 3

As a result, the minimum value of P P is 7 \boxed{7} , and the partial fractions can be completed as the following:

x 2 − 7 ( x − 2 ) ( x − 3 ) ( x − 5 ) = − 1 x − 2 − 1 x − 3 + 3 x − 5 \dfrac{x^2 - 7}{(x - 2)(x - 3)(x - 5)} = \dfrac{-1}{x - 2} - \dfrac{1}{x - 3} + \dfrac{3}{x - 5}

11 Helpful 2 Interesting 2 Brilliant 0 Confused

I also did the same way .. but what if that prime number was large . I mean by substituting how many values could we check ? So isn't there any other method for it ?

Anurag Pandey - 4 years, 9 months ago

this was just trial and error method but is there any efficient method to solve this problem

abhishek alva - 4 years, 8 months ago

Just like to add this: once you see (4-p)/3, (9-p)/-2, and (25-p)/6 we can see that (treat = as congruent plz) p=1 mod 3, p=1 mod 2, and p= 1 mod 6. Furthermore P is prime. The most obvious first value is 3 (from 1+2). However this doesnt work as 21/6 isn’t an integer. Then we move to 7 (the first residue of 1 mod 6) and see that this works and is also the smallest value.

Mr. Krabs - 2 months, 1 week ago

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