-4
$\begingroup$

So $81=9\times 9$ , $225=9\times 25$, but numbers like 9, 25, 49, 121, 169... the prime squares can not be written as a product of squares. If so, I may be able to make a formula for the $n$-th prime square which would be the $n$-th prime!

  • 0
    1 is a square number – 2017-01-07
  • 0
    I suppose the exception is 1 – 2017-01-07
  • 0
    What is a "prime square" in this case? Also, the wording is very unclear in general... – 2017-01-07
  • 0
    Squaring is multiplicative, $(ab)^2=a^2b^2$. That means $n^2$ is a product of other squares $a^2$ and $b^2$ if and only if $n$ is a product of other numbers $a$ and $b$. – 2017-01-07
  • 0
    Of course! Any number that is not a prime is $n=mk$ so $n^2 = m^2k^2$ is a product of squares. So for $n^2$ to not be a product of squares (other than $1 = 1^2$) then $n$ must be prime. – 2017-01-07
  • 0
    Though your observation is valid, it doesn't seem to simply the task of finding primes - you "may" be able to - but do you have a good idea? – 2017-01-07
  • 0
    For s=(2n+1)^2, n>0 and an element of integers. My thinking is 'how to distinguish a prime number'. Every odd number squared can be written as the product of two squares, except for prime numbers squared, so that the n-th square which cannot be represented as a product of two existing squares (values given by s(n)) must be a prime number squared. – 2017-01-07

1 Answers 1

1

Consider the number $n^2$. If $n$ is not a prime we can write it as $n=kd$ with $k,d>1$.

It follows that $n^2=k^2 d^2$.

So yes, if a square does not have a proper square divisors then that number is the square of a prime.