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If a prime number on division by $4$ gives a remainder of $1,$ then that number can be expressed as

  1. sum of squares of two natural numbers
  2. sum of cubes of two natural numbers
  3. sum of square roots of two natural numbers
  4. sum of cube roots of two natural numbers
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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.

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we can express them as sum of square roots of 2 natural numbers definitely.

eg: 17,29,41 etc..

17 = 9 +8 =Sq root of 81 + Sq root of 64

like that we can express.

Option C is answer i guess
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x mod 4=1 ..13 can be one such number and it is also prime ..can be expressed as ..2 2+32=>13..

so option A seems to be the correct one 

can be applied on 17-> 42+12... and so on..

Answer:

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