# How many whole number terms does $(\sqrt[11]5+\sqrt[5]11)^{2015}$ have?

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

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

• 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 $55$. – Jyrki Lahtonen Jun 9 at 14:36
• It usually looks better, I find, if you write fractional exponents as $k/11$ instead of ${k\over11}$ – saulspatz Jun 9 at 14:54

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.

• How do you explain the different form, is it incorrect – Milan Jun 9 at 14:45
• 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. – elescararriba Jun 9 at 14:48
• 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. – elescararriba Jun 9 at 15:03

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?

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.

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)