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Find all possible positive integers $n$ such that $3^{n-1} + 5^{n-1} \mid 3^n + 5^n $.

I don't know how to start with. Any hint or full solution will be helpful.

4 Answers 4

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Let's suppose that $3^{n-1}+5^{n-1}\mid 3^n+5^n$, so there is some positive integer $k$ such that $3^n+5^n=k(3^{n-1}+5^{n-1})$. Now, if $k\ge 5$ we have $$k(3^{n-1}+5^{n-1})\ge 5(3^{n-1}+5^{n-1})=5\cdot 3^{n-1}+5\cdot 5^{n-1}>3^n+5^n.$$

This means that $k\le 4$. On the other hand, $3^n+5^n=3\cdot 3^{n-1}+5\cdot 5^{n-1}>3(3^{n-1}+5^{n-1})$, then $k\ge 4$ and thus we deduce that $k=4$. In this case we have $3^n+5^n=4(3^{n-1}+5^{n-1})$, which gives us the equation $5^{n-1}=3^{n-1}$, but if $n>1$ this equation is impossible.

Hence $n=1$, and we can easily check that $2\mid 8$.

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Hint: $A:=3^{n-1}+5^{n-1}$, and $B:=3^n+5^n$.
Compare $3A$ and $5A$ to $B$ to get the possible values of the integer $B/A$.

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    dont understand :( Still cant solve – 2017-01-22
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    That's a hint to start. If $A\,|\,B$ then what can the integer $B/A$ be? Can it be $3,4,5$? – 2017-01-22
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    I didn't get what you mean by " Put in $3 \cdot A$ between" – 2017-01-22
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    I mean to compare $3A$ to $B$, and then you will see $A<3A$5A$ to $B$. – 2017-01-22
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    I think you mean $3A > \frac{B}{2}$ and $5A>B$, so if $A|B$, either $2A = B$, or $4A=B$, which are obviously false for $n>1$. – 2017-01-22
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Let $d$ be a positive integer such that $d$ divides both $3^{n-1}+5^{n-1}$ and $3^{n}+5^{n}$.

then d divides $5\cdot (3^{n-1}+5^{n-1})-( 3^{n}+5^{n} )= 3^{n-1}\cdot 2$, and

$3\cdot(3^{n-1}+5^{n-1})-( 3^{n}+5^{n} )= -5^{n-1}\cdot 2$

Thus, as $\gcd(5,3)=1$, it follows that $d$ is either $1$ ou $2$.

Since $3^{n-1}+5^{n-1}$ divides $3^{n}+5^{n}$, we must have $d=3^{n-1}+5^{n-1}=2$, which gives $n=1$.

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If $n= 1$ then we have $3^{n-1} + 5^{n-1} = 1+ 1 = 2|8 = 3^{n}+5^{n}$. So that's one solution.

Assume $n > 1$.

$3^n + 5^n = 3*3^{n-1} + 5*5^{n-1} = 3*3^{n-1} + 3*5^{n-1} + 2*5^{n-1} = 3(3^{n-1} + 5^{n-1}) + 2*5^{n-1}$.

So $3^{n-1} + 5^{n-1}|2*5^{n-1}$

If $p$ is prime and $p|3^{n-1} + 5^{n-1}$ then $p|2*5^{n-1}$ then $p = 5$ or $p= 2$

If $p = 5$ then $p\not |3^{n-1} + 5^{n-1}$. So $p = 2$ and $3^{n-1} + 5^{n-1} = 2^k$ but $2^k| 2*5^{n-1}$ means $k = 1$ so $3^{n-1} + 5^{n-1} = 2$. Which can only happen if $n = 1$.

So $n =1$ is only solution.