Assume that $p,q,r,s$ are prime. If $$p^3q^2=r^2s^3,$$ then according to the fundamental theorem of arithmetic:

(i) $ \ q=s \ $

(ii) $ \ q >r \ $

(iii) $ \ p^3=s^2 \ $

(iv) $ \ p^3<s^2 \ $


We have

$ p^3q^2=r^2s^3 \ \Rightarrow pppqq=rrsss \ ...........(1) $

If $ \ p=r=s \ $ , then from $ \ (1) $ , we get

$ rrsqq=rrsss \\ \Rightarrow qq=ss, \ \ (by \ \ left \ \ cancel) \\ \Rightarrow q=s $ ,

Thus option (i) is true.

what about the other options ?

  • 3
    $\begingroup$ $p= s, q = r$ none of the options are correct. $\endgroup$ – Doug M May 24 '18 at 19:20
  • 1
    $\begingroup$ @DougM The (i) is not true in general but it is compatible with the condition given by FTA. $\endgroup$ – gimusi May 24 '18 at 19:23
  • $\begingroup$ @DougM and also (iii) $\endgroup$ – gimusi May 24 '18 at 19:27
  • $\begingroup$ @gimusi The way the problem is posed, we look for options that are true in general. (ii), (iii), (iv) are clearly incompatible, but (i) is still only compatible and not an implication $\endgroup$ – Hagen von Eitzen May 24 '18 at 19:29
  • $\begingroup$ @gimusi For which values is iii true? $\endgroup$ – miracle173 May 24 '18 at 19:29

$N= p^3q^2 =r^2s^3$.

If $p$ and $q$ are different primes than $N$ has precisely two prime factors one to the precise power of three and the other to the precise power of two and so $p = s$ and $q = r$.

If $p$ and $q$ are the same prime then $N$ has only one prime factor and it is to the fifth power so $p = q = r = s$.


i) $q=s$. That could happen but is doesn't have to. If it does though then $p=q=r=s$. If $p \ne q$ then $s = p\ne q$.

ii) $q > r$. That's impossible as $q = r$ under all cases. However the FTA says nothing about the relative sizes of different primes. We can't say anything about whether $q$ is greater less or equal to $s=p$.

iii) $p^3 = s^2$ that'd be impossible for any two primes under any circumstances. Each number has a unique prime factorization and so if it has only one prime factor it must be to a specific power. It can't be one prime to the third and another prime squared.

iv) $p^3 < s^2$. Under all cases $p = s$ and as $p = s > 1$ then $p^3 > p^2 = s^2$ so this is impossible. Again if the primes were not equal we would no nothing about relative sizes.


i) is compatible but certainly not true, in the sense that it must be or is even likely to be.

ii)-iv) are all impossible.

So answer: none of them.


According to the fundamental theorem of arithmetic we have that

$$ p^3q^2=r^2s^3 \implies p=s \,\land \, q=r$$

and then the implication is false for all the options but the options (i) is compatible with that for the particular case $p=s=q=r$ even if not true in general for the given condition.

  • $\begingroup$ All options are true when not taken in general $\endgroup$ – Love Invariants May 24 '18 at 19:23
  • $\begingroup$ @LoveInvariants $q>r$ is always false and also $p^3<s^2$ but (i) and (iii) can be true if p=q=r=s $\endgroup$ – gimusi May 24 '18 at 19:25
  • $\begingroup$ Also none of the options are correct. But we always talk in general $\endgroup$ – Love Invariants May 24 '18 at 19:28
  • $\begingroup$ @LoveInvariants I agree that we can only conclude that p=s and q=r but options (i) and (iii) can be true also for the particular case p=s=q=r otherwise (ii) and (iv) are always false. $\endgroup$ – gimusi May 24 '18 at 19:30
  • $\begingroup$ if at least one is true then i and only i is true $\endgroup$ – miracle173 May 24 '18 at 19:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy