This question was uploaded on 15/10/21 on social media accounts.

Given that a and b are prime numbers such that (a^b + b^a) is also a prime number. Then find all pairs of (a, b).

Solution:

Given that a, b are prime numbers. There is only one even prime number '2' and the rest of the others are odd.

If both even: (a^b + b^a) will be even and greater than 2 (Not a prime number)

If both odd: (a^b + b^a) will be even and greater than 2 (Not a prime number)

So here one must be even and the other must be odd, but we know there is only one even prime number. Let a is even means a = 2.

Now, (a^b + b^a) = (2^b + b^2).

Now, as we know b is odd.

Let,

\[b = 2n + 1\]

\[2^b = 2^{2n-1} = 2\cdot4^n = 2\cdot(3+1)^n\]

\[(3+1)^n=1⠀(mod⠀3)\]

Means remainder is 1 when divided by 3

\[\Rightarrow 2\cdot(3+1)^n=2⠀(mod⠀3)\]

\[\Rightarrow 2^b = 2⠀(mod⠀3)---(1)\]

Means remainder is 2 when divides by 3

Now, as let assume b is odd and not a multiple of 3 (Not a multiple of any prime number but we are taking only 3. You will understand why we take only 3 in this case.)

Let,

\[b = 6m \pm 1\]

\[b^2 = (6m \pm 1)^2=36m\pm12m+1\]

\[\Rightarrow b^2 =3(12m\pm4m)+1\]

\[\Rightarrow b^2 = 1⠀(mod⠀3)---(2)\]

From both equations, (2^b + b^2) is divided by 3.

That means for every odd number (other than multiples of 3), (2^b + b^2) is divided by 3 means, not a prime number. And for all multiples of 3, b is not a prime number since it is multiple of 3 so it can be divided by 3.

That means for any odd number greater than 3, (a^b + b^a) is not a prime number.

Let's check for number 3:

\[a^b + b^a = 2^3 + 3^2 = 17\]

17 is a prime number so answers are: (a, b) = (2, 3), (3, 2)

Puzzle based on Algebra, Equation.