Number of distinct terms in the expansion of $\big(x+\frac{1}{x}+x^2+\frac{1}{x^2}\big)^{15}$ 
Number of distinct terms in the expansion of $\bigg(x+\dfrac{1}{x}+x^2+\dfrac{1}{x^2}\bigg)^{15}$ is equal to ?

We can write the above as,
$$ \bigg(x+\dfrac{1}{x}+x^2+\dfrac{1}{x^2}\bigg)^{15} = \dfrac{1}{x^{30}}(1+x+x^3+x^4)^{15} $$
Now my teacher says that we can expand the polynomial $(1+x+x^3+x^4)^{15} $ as,
$$ (1+x+x^3+x^4)^{15} = a_0 + a_1x + a_2x^2 + a_3x^3.....a_{60}x^{60} $$ 
Hence, as each term is divided by $x^{30}$, the number of distinct terms would be equal to 61. 
But my question is how do we know that the expansion of the polynomial $(1+x+x^3+x^4)^{15} $  will contain all powers of $x$ from $x^0$ to $x^{60}$ ?
 A: A general term (without the coefficient) of the polynomial $(1+x+x^3+x^4)^{15}$ will look like
$$x^{a+3b+4c} \qquad \text{ where } 0 \leq a,b,c, \leq 15.$$
So your question is now to show that $a+3b+4c$ can take on all integer values between $0$ and $60$. 
For numbers in $0 \to 15$ let $a$ vary and have $b=c=0$.
Now try doing the same for other ranges to convince yourself that all powers will be obtained for the right combinations of $a,b$ and $c$.
A: Your expression is equal to
$$
\frac1{x^{30}}(x^3+1)^{15}(x+1)^{15}
$$
$(x^3+1)^{15}$ contains $16$ terms, each containing a power of $x^3$ with a positive coefficient. Since $(x+1)^{15}$ contains $16$ terms of consecutive powers of $x$ with positive coefficients, the product will fill in the gaps in $(x^3+1)^{15}$. Thus, in your expression, there are non-zero terms from $x^{30}$ to $x^{-30}$.
In fact, the coefficient of $x^n$ is the number of solutions to $4a+3b+c=n$ where $a+b+c\le15$ and $a,b,c\ge0$. Thus, there is always at least one solution for $0\le n\le60$ and no solutions for other values of $n$.
That means $61$ distinct terms.
