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

 
 Jun 11, 2024
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From \(x^2-mx+n=0\), we get k+t=m and kt=m. Since n is prime, one of k and t is n and the other is 1. K>t, so k=n and t=1. Then m=n+1. m is also prime, so we have two consecutive integers that are prime. Since one of every two consecutive integers is even, and the only even prime is 2, we must have n=2 and m=3. Therefore, \(m^n+n^m+k^t+t^k= 3^2+2^3+2^1+1^2=9+8+2+1=\boxed{20}\).

 Jun 11, 2024

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