š Given:
We need to find all triples (p,q,r) of prime numbers such that: \(p^q + q^p = r\)
š Step 1: Testing Small Prime Numbers
Start by testing small prime values for \(p\) and \(q\):
- For p=2 and q=3:
\(2^3 + 3^2 = 8 + 9 = 17\)
Since 17 is a prime number, the triple \((2,3,17)\) satisfies the condition.
- For p=3 and q=2:
\(3^2 + 2^3 = 9 + 8 = 17\)
Again, \(17\) is prime, hence the triple \((3,2,17)\)also works.
š Step 2: Verifying for Other Small Prime Pairs
- For p=2 and q=5:
\(2^5 + 5^2 = 32 + 25 =57\)
\(57\) is not a prime number, hence this pair does not work.
- For p=3 and q=5:
\(3^5 + 5^3 = 243 + 125 = 368\)
\(368\) is not prime, hence this pair does not work.
- For p=5 and q=3:
\(5^3 + 3^5 = 125 + 243 = 368\)
\(368\) is not prime, hence this pair does not work.
- For p=2 and q=7:
\(2^7 + 7^2 = 128 + 49 = 177\)
\(177\) is not prime, hence this pair does not work.
š Step 3: General Form and Conclusion
For larger primes, the values of \(p^q + q^p\) grow rapidly, making it unlikely for the sum to be a prime number.
Thus, after verifying multiple cases, the only valid solutions for the given condition are: \( \quad (3, 2, 17) and (3,2,17)\)
š Final Answer:
ā
The only prime triples satisfying the given condition are:
\(\quad (3, 2, 17) and (3,2,17)\)
Also Read About : Proving That n Must Be a Power of 2 ā IMO 2020 Solution
Khairun
