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If $n$ is a natural number and $n$ is a $4th$ power and a $5th$ power prove it is a $20th$ power. (Hint: Use fundamental theorem of arithmetic).

I can't do this problem and am looking for advice.

I know you can express $n$ as two different numbers $x,y$ say one to the $4th$ power the other to the $5th$ power. And each of these can be expressed as the product of increasing prime numbers.

I somehow need to get that there exists a number such that when you put it to the $20th$ power you get $x$.

It looks like I'm very close in my writings but I can't seem to make that vital last step.

I'm guessing it has something to do with manipulating the different equations for $x$.

Any help?

Christof
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3 Answers3

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Hint: $n=\prod p_i^{\alpha_i}$.

Since $n=x^4=y^5$, what can we say about the factorizations of $x^4,y^5$, and consequently about $\alpha_i$?

Some Math Student
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  • $(\prod p_i^{\alpha_i})^{\frac{1}{4}}=x$ and $(\prod p_i^{\alpha_i})^{\frac{1}{5}}=y$. – Christof May 22 '15 at 17:11
  • Hmmmm so can we take the product $xy$ and get $(\prod p_i^{\alpha_i})^{\frac{1}{20}}=t$. So this is the number we want? – Christof May 22 '15 at 17:13
  • Either that, when you need to state that $4|\alpha_i$ else $x$ is not an integer. Note, though, that exponents add when multiplying, so your property for $t$ doesn't hold. Else you could simply write $x^4=\prod p_i^{4\beta_i}$, and the fundamental theorem tells you that $4\beta_i=\alpha_i$, and then... – Some Math Student May 22 '15 at 17:21
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Consider any prime factor $p$ of $n$. $n$ will be divisible by some positive power $k$ of $p$, so $p^k \mid n$, taking the maximum $k$.

Since $n$ is a fourth power, we know that $p^{k/4} \mid \sqrt[4]n$ - so $4 \mid k$. And since $n$ is a fifth power, we know that $p^{k/5} \mid \sqrt[5]n$ - so $5 \mid k$.

Therefore we also know that $\text{lcm}(4,5) \mid n$, and since $\gcd(4,5) = 1$, that means that $20 \mid k$.

Since this is true for any prime divisor on $n$, all the prime power exponents in the prime decomposition of $n$ will be divisble by $20$ and hence $n$ is a $20^{th}$ power.

Joffan
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Note that $n^5=x^{20}$ and $n^4=y^{20}$ so that $$n=\frac {n^5}{n^4}=\left (\frac xy\right)^{20}$$

And you may know a way to prove that if $q$ is a rational number with $q^r=n$ where $r$ and $n$ are positive integers, then $q$ is an integer. Or it is easy to prove this special case.

This doesn't use the fundamental theorem of arithmetic directly, but it is there in the background of the second paragraph. And I've left that key part of the proof to you. But I thought showing how $n$ can be directly expressed as the twentieth power of something might be of interest.

Mark Bennet
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