# Do there exist primes a, b, c satisfying a^2+b^3=c^4?

Feb 10, 2017

No

#### Explanation:

Suppose ${a}^{2} + {b}^{3} = {c}^{4}$ for some prime numbers $a , b , c$.

Note that $a = b = c = 2$ does not work, meaning for both sides of the equation to have the same parity, one of $a , b , c$ must be $2$, and the other two must be odd primes.

Next, note that we may subtract ${a}^{2}$ from each side of the equation to get

${b}^{3} = {c}^{4} - {a}^{2} = \left({c}^{2} + a\right) \left({c}^{2} - a\right)$

$\implies {b}^{2} = {c}^{2} + a$ and $b = {c}^{2} - a$

(as ${c}^{2} + a > {c}^{2} - a$ and $b$ is prime)

$\implies {\left({c}^{2} - a\right)}^{2} = {c}^{2} + a$

We now consider three cases.

Case 1: $a = 2$

$\implies {\left({c}^{2} - 2\right)}^{2} = {c}^{2} + 2$

$\implies {c}^{4} - 4 {c}^{2} + 4 = {c}^{2} + 2$

$\implies {c}^{4} - 5 {c}^{2} + 2 = 0$

$\implies {c}^{2} = \frac{5 \pm \sqrt{17}}{2}$

$\implies c \notin \mathbb{N}$

but this contradicts the premise that $c$ is a prime.

Case 2: $b = 2$

$\implies \left\{\begin{matrix}{c}^{2} - a = 2 \\ {c}^{2} + a = 4\end{matrix}\right.$

$\implies \left({c}^{2} - a\right) + \left({c}^{2} + a\right) = 2 + 4$

$\implies 2 {c}^{2} = 6$

$\implies {c}^{2} = 3$

$\implies c \ne \mathbb{N}$

again contradicting the premise that $c$ is prime.

Case 3: $c = 2$

$\implies {a}^{2} + {b}^{3} = 16$

If $a$ and $b$ are both odd primes, then the least the left hand side can attain is when both $a$ and $b$ are the least odd prime, i.e. $a = b = 3$, giving ${a}^{2} + {b}^{3} \ge {3}^{2} + {3}^{3} > 16$, a contradiction.

As each case leads to a contradiction, there are no three primes $a , b , c$ satisfying ${a}^{2} + {b}^{3} = {c}^{4}$