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3 Answers

Best answer
7 7 votes

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.

• selected by
0 0 votes
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
0 0 votes

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..

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