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Proving one way is very simple

If n is a perfect square then

n = $a^2$

$n^7$ = $a^{14}$

$n^7$ = $a^7$($a^7$) which is obviously a perfect square for some integer a.

It's the if $n^7$ is a perfect square then n is a perfect square part that is giving me trouble. I have already tried to get hints from others and the one hint I was given was to write $n^7$ as a product of primes, but I have no clue where to go from there. I believe it can be done using modulus but I honestly have no idea.

Any hints would help thanks.

randomuser
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    This is entirely uniqueness of the prime factorization. What can be said about the exponents in the prime factorization of a square? – Will Jagy Mar 05 '20 at 04:17
  • By the linked post $\large , n^7 = a^2,\Rightarrow, n = b^2,\ a = b^7,$ for $\large ,b\in\Bbb Z\ \ \ $ – Bill Dubuque Mar 05 '20 at 04:53

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Maybe it is helpful to observe that $n^6$ is a perfect square and $n=\frac{n^7}{n^6}$.

lizhitai
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    To reduce it to showing that if a whole number has a rational square root then it must be a perfect square? – Jonas Meyer Mar 05 '20 at 04:30
  • @JonasMeyer It might be sufficient to show that if $a$ divides $b$ and both of them are squares, then their quotient is also a square. – lizhitai Mar 05 '20 at 04:49
  • Is that any different? If $n = (\frac{x}{y})^2$, then $y^2$ divides $x^2$. – Jonas Meyer Mar 05 '20 at 05:04
  • @Jonas The hint is actually special case of various proofs in the linked dupe. It is clearer in additive form, viz. $ $if $,2\mid 7n,$ then $,2\mid 6n,\Rightarrow, 2\mid n = 7n-6n.\ $ One of my answers in the dupe compares the multiplicative and additive forms. – Bill Dubuque Mar 05 '20 at 05:13