The two positive integer solutions of the equation $x^2 - mx + n = 0$ are $k$ and $t$, where $m$ and $n$ are both prime numbers and $k > t$. What is the value of $m^n + n^m + k^t + t^k$?

Question

contestada

Respuesta :

DeanR DeanR · Answer 1 · 2019-12-24T20:58:53+01:00

Since k and t are roots of x² - mx + n = 0,

0 = (x - k)(x - t) = x² - (k+t)x + kt = x² - mx + n

Equating respective coefficients,

m = k + t and n = kt

If n is prime we have one of k and t must be a unit ±1. Since we're told they're positive, the smallest must be +1, so we conclude

t = 1

So

n = kt = k

We're told m is prime

m = k + t = n + 1

m and n are consecutive and prime. The only consecutive primes are 2 and 3. Our solution is

k = 2, n = 2, m = 3, t = 1

We're inexplicably asked for the symmetric quantity

[tex]m^n + n^m + k^t + t^k[/tex]

We compute

3² + 2³ + 2 + 1² = 9 + 8 + 2 + 1 = 20

Answer: 20