Theorem$:$ If $p$ is a prime number which has remainder $1$ when divided by $4,$ then $p$ can be written as a sum of two squares.
(Reference: http://math.arizona.edu/~wmc/Courses/323044/Lecture4.pdf )
Example$:$
- $5 = 1 + 4$
- $13 = 9 + 4$
- $17 = 16 + 1$
- $29 = 25 + 4$
- $37 = 36 + 1$
- $41 = 25 + 16$
- $53 = 49 + 4$
- $61 = 36 + 25$
- $73 = 64 + 9$
- and so on.
So, option $(A)$ is correct.
Take option $(B)$ sum of cubes of two natural numbers.
Counter example$:\ 5$ cannot represent as cubes of two natural numbers $(1^{3}=1,2^{3}=8\implies 1 + 8 = 9\neq 5).$
So, option $(B)$ is not correct.
Take option $(C)$ sum of square roots of two natural numbers
Let $p = 4n + 1, n \geq 0$ and $p$ is prime.
The smallest such $p$ is $5.$
So, any $p$ can be written as $ p = p_1 + p_2$ where $p_1$ and $p_2$ are natural numbers. Now, existence of $p_1^2$ and $p_2^2$ makes option C correct.
For option D, instead of $p_1^2$ and $p_2^2$ in above explanation we just need to change to $p_1^3$ and $p_2^3.$
Correct options $: \ (A), \ (C), \ (D).$
It is better to pick option A here, because Options C and D seem to be given by mistake. Nowadays at least in GATE this will cause Marks to All during debate.