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$$(\sqrt[5]11+\sqrt[11]5)^{2015} $$

It is unnecesary to write binomial coefficents $$11^{\frac k{5}}5^{\frac {2015-k}{11}} $$ This is the general term of the sum , $k$ goes from $0$ to $2015$. For a term to be a whole number it means that : $$5\;|\;k\quad \wedge\; 11\;|\;2015-k $$ $2013$ is divisble by $11$ , so $11\;|\;2013+2-k=11\;|\;2-k$

$k$ should be divisible by $5$ and $k-2$ divisible by $11$. In the solutions it says that from this it follows that k should be of the form: $k=55l+35 $. I dont understand this.

On the other hand if we write the binomial expansion in different order things are simpler $$5^{\frac k{11}}11^{\frac {2015-k}{5}} $$ $k$ should be divisible by $ 11$ and by $5$ which makes k of the form $k=55l$.

These forms are different I don't know how to explain it. They both give the correct solution. There are 37 whole number terms ( $2015=55*36+35$, and $k=0 $)

Edit: if someone could make the exponents look bigger

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  • $\begingroup$ Five and eleven are coprime, so the Chinese remainder theorem is all you need to conclude that the solutions form a single residue class modulo $55$. $\endgroup$ – Jyrki Lahtonen Jun 9 at 14:36
  • $\begingroup$ It usually looks better, I find, if you write fractional exponents as $k/11$ instead of ${k\over11}$ $\endgroup$ – saulspatz Jun 9 at 14:54
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ok, $11$ divides $k-2$ means that there is an $m$ such that $k=11m+2$. But $m$ can be of the form $5l,5l+1,5l+2,5l+3$ or $5l+4$. Substituting you get that $k$ is of the form $55l+2,55l+13,55l+24,55l+35$ or $55l+46$. So then it's not difficult to understand why $55l+35$ is the only possible solution. If you think about it and don't find why just ask.

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  • $\begingroup$ How do you explain the different form, is it incorrect $\endgroup$ – Milan Jun 9 at 14:45
  • $\begingroup$ no, the other possibilities are not possible because $5$ divides $k$. Now what you have to do is to prove that the $k$ of the form $55l+35$ are solutions. $\endgroup$ – elescararriba Jun 9 at 14:48
  • $\begingroup$ Ah sorry now I know your problem. Yes, the problem is that your $k$ is not the same $k$ as in the book. If $k$ is the one of the book and $k'$ is yours, you get that $k'$ is of the form $55l'$ but in you case $k'=2015-k$, but $2015=55\cdot 36+35$ so $k=2015-k'=55(36-l')+35$. So the $l$ of the book is $36-l'$. So both are right but you call things in a different way. $\endgroup$ – elescararriba Jun 9 at 15:03
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There really is no reason that the formulas should be the same. In the given solution, $k/5$ is the exponent of $11$. In your solution, $k/11$ is the exponent of $5$. It just happens that you've used the same letter. If you wrote the exponent of $5$ as $j/11$ you wouldn't expect that $j=k$, now would you?

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The forms aren't different -- let's write the first one as $$11^\frac{k}{5}5^\frac{2015-k}{11}$$ and the second as $$5^\frac{j}{11}5^\frac{2015-j}{5}$$ As you said, the solutions for $k$ are $k\equiv35\mod55$, and for $j$ they are $j\equiv0\mod55$

How are $k$ and $j$ related? Well, $j=2015-k$. But if we look carefully, we see that $k\equiv35\mod55$ and $j=2015-k$ together imply that $j\equiv0\mod55$ so in fact the two equations in $j$ and $k$ are equivalent.

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Given $5\mid k$ and $11\mid 2015-k$.

$5\mid k \quad\Longrightarrow\quad k\equiv 0\equiv\pmod{5}$

$11\mid 2015-k=11\mid 2-k \quad\Longrightarrow\quad k\equiv 2\equiv\pmod{11}$

Here you need to apply Chinese remainder theorem:

$k\quad\equiv\quad Chinese(0\equiv\pmod{5},2\equiv\pmod{11})\quad \equiv \quad35\equiv\pmod{55}$

gp-function:

? chinese(Mod(0,5),Mod(2,11))
%1 = Mod(35, 55)
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