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I might have a silly questions but I am new to math so don't judge to hard, is there a proof that there is no integer for which $a^n = b^2$ where $n>2$ ?

PS The question excludes the case where (a=c^2$) (b=c^n$). Example( 4 and 8, 16 should be excluded) also solutions a=b=0 or a=b=1 also to be excluded.

Hivaga
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    $9^3=27^2$. In general, $(a^2)^n=(a^n)^2$. – lulu Jun 06 '21 at 23:55
  • If $a=c^2$ and $b=c^n$, for any natural number $c$, then in fact $a^n=b^2=c^{2n}$. For instance, $4^3=8^2$. – mjqxxxx Jun 06 '21 at 23:58
  • Tnx, but I am talking more about the case where a and b are not equal to c^2 or c^n – Hivaga Jun 07 '21 at 00:01
  • @Hivaga you should edit the question to reflect your intentions if you want to exclude the case that satisfies equation. – CyclotomicField Jun 07 '21 at 00:04
  • @MichaelMorrow that is a good point :) – Hivaga Jun 07 '21 at 00:10
  • Suggestion for the question you appear to be asking: Consider the prime factorizations of $a$ and $b$. – Andrew D. Hwang Jun 07 '21 at 00:38
  • "The question excludes the case where (a=c^2)(b=cn). Example( 4 and 8, 16 should be excluded) also solutions a=b=0 or a=b=1 also to be excluded." You should really clarify your actual question. The answer to the question is "no" because it isn't true. If $a = c^k$ and $b=c$ and $n = 2k$ is, of course, a counter example. You can't say "but that doesn't count because..." because you didn't clarify what you were actually asking form. – fleablood Jun 07 '21 at 02:31
  • By the dupe if $n$ is odd then $,a = c^n, b = c^2,$ for an integer $,c,,$ which is excluded. Thus $n$ is even, where obviously it is solvable: let $,a = b^{n/2}\ \ $ – Bill Dubuque Jun 07 '21 at 08:28

2 Answers2

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Suppose that $a$ and $b$ are positive integers with $a^n = b^2$, where $n> 2$ is an integer. Then $a = \sqrt[n]{b^2} = [\sqrt[n]{b}]^2$. You want to know all the ways this can happen.

If $a$ is a perfect square, say $a = c^2$ for a positive integer $c$, then you can take $b = c^n$, and then $a^n = c^{2n} = b^2$.

If $a$ is not a perfect square, then this situation can only happen when $n$ is even. Indeed, if $a = p_1^{e_1} \cdots p_s^{e_s}$ and $b = p_1^{f_1} \cdots p_s^{f_s}$ are the prime factorizations of $a$ and $b$ (with $e_i, f_i \geq 0$), then $a^n = b^2$ implies that $ne_i = 2f_i$ for each $i$. The assumption that $a$ is not a perfect square is equivalent to the claim that some $e_i$ is an odd number. For such $i$, the equation $ne_i = 2f_i$ implies that $n$ is even, as claimed.

So for $n$ even, every positive integer $a$ admits a corresponding positive integer $b$ such that $a^n = b^2$, since you can just take $b = a^{n/2}$ (an integer!).

For example, if $n = 4$, you can take for $(a,b)$ the pairs

$$(1,1); (2^4, 4^2); (3^4,9^2); (4^4,16^2); ...$$

D_S
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There is a well known fact that $\sqrt[k]{m}$ (where $m$ is an integer) is an integer only if $m$ is an integer to a $k$ power.

So if $a^2 = b^n$ then either $b$ is a perfect square or $n$ is even.

If $b = c^2$ for some integer $c$ then $a= c^n$ and $a$ is an integer to the $n$th power.

And if $n = 2k$ then $a = b^k$ and $a$ is a perfect $k$ power. And that's the only way it is possible (although that is quite possible).

I'm not sure what you are looking for but that pretty much exhausts everything.

fleablood
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