Prime Quadratic Equation

Algebra Level 2

p p and q q are primes greater than 10 where 2 p − 1 = q { 2 }^{ p }-1=q .

Find the minimum possible value of p p and q q and choose the quadratic equation where p 2 {p}^{2} and q q are roots.

x 2 + 8360 x + 1384279 = 0 { x }^{ 2 }+8360x+1384279=0 x 2 − 8360 x − 1384279 = 0 { x }^{ 2 }-8360x-1384279=0 x 2 − 8360 x + 1384279 = 0 { x }^{ 2 }-8360x+1384279=0 x 2 + 8360 x − 1384279 = 0 { x }^{ 2 }+8360x-1384279=0

Victor Paes Plinio by Victor Paes Plinio , 21, Brazil

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2 solutions

Actually you don't have to find the solution to choose the correct answer, because obviously p 2 p^2 and q q are both positive, and the only option that has positive roots is x 2 − 8360 x + 1384279 = 0 x^2-8360x+1384279=0 .

But we see that 2 13 − 1 = 8191 2^{13}-1=8191 is prime, hence p = 13 p=13 and q = 8191 q=8191 .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

THats the way i used too ...........Bingo buddy !

Apoorv Padghan - 6 years, 10 months ago

The question is absolutely wrong. 169 is not a prime.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

I say p p is prime 13 13 , but I don't say p 2 {p}^{2} is prime, but the root from the equation.

Victor Paes Plinio - 6 years, 11 months ago

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