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

Question

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

Solution: 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 $n=1$

The above solution is from this answer. Could someone please explain a few things:

• why was the value of $k\le4$ and $k\ge4$ picked? where did they get 4?
• why does $3\cdot 3^{n-1}+5\cdot 5^{n-1}>3(3^{n-1}+5^{n-1})$ mean that $k\ge4$ ?
• Why does $5\cdot 3^{n-1}+5\cdot 5^{n-1}>3^n+5^n.$ mean that $k\le4$ ?

Answer 1

The line $$k(3^{n-1}+5^{n-1})\ge 5(3^{n-1}+5^{n-1})=5\cdot 3^{n-1}+5\cdot 5^{n-1}>3^n+5^n$$ makes a contradiction, since $k(3^{n-1}+5^{n-1})=3^n+5^n$, and $3^n+5^n>3^n+5^n$ is obviously false. Since they assumed $k\geq 5$, that must be false; thus, we conclude $k\leq 4$.

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Then they say $$3^n+5^n>3(3^{n-1}+5^{n-1})$$ which you could rewrite to $$3^n+5^n=k(3^{n-1}+5^{n-1}>3(3^{n-1}+5^{n-1})$$ thus $k>3$, and since $k\leq 4$, we know $k=4$.

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They're basically just first eliminating all numbers greater than four, and then all integers smaller than four, to conclude $k=4$ (which they eliminate too).

Answer 2

Perhaps an easier way to look at it: the equation $$3^n+5^n=k(3^{n-1}+5^{n-1})$$ can be rewritten $$5^{n-1}(5-k)=3^{n-1}(k-3)\ .$$ Therefore $5-k$ and $k-3$ must have the same sign. They can't both be negative as then $k$ would be simultaneously less than $3$ and greater than $5$, and they clearly can't both be zero, so they must both be positive. Hence $$k<5\quad\hbox{and}\quad k>3\ ,$$ so $k=4$. This makes our second equation $$5^{n-1}=3^{n-1}\ ,$$ and this is true if and only if $n-1=0$.