\section*{Problem 1 - 151221}

a) Investigate whether there exist natural numbers \(n\) such that in the expansion formed according to the binomial theorem

\[
(a+b)^{n}=c_{0} a^{n}+c_{1} a^{n-1} \cdot b+c_{2} a^{n-2} \cdot b^{2}+\ldots+c_{n} b^{n}
\]

the coefficients \(c_{0}, c_{1}, c_{2}\) have the sum \(c_{0}+c_{1}+c_{2}=79\). If such numbers \(n\) exist, determine them.

b) Investigate whether there exist natural numbers \(n\) such that from (1) by substituting \(a = x^{2}, b = \frac{1}{x}\) an expansion

\[
\left(x^{2}+\frac{1}{x}\right)^{n}=c_{0} x^{k_{0}}+c_{1} x^{k_{1}}+c_{2} x^{k_{2}}+\ldots+c_{n} x^{k_{n}}
\]

is formed, in which one of the exponents has the value \(k_{i}=0\), i.e., in which a term free of \(x\) appears. If such numbers exist, determine them.

b) Determine all natural numbers \(n\) that satisfy both the conditions specified in a) and b).
}

a) For fixed \(n\), we have \(c_{0}=\binom{n}{0}=1\), \(c_{1}=\binom{n}{1}=n\), and \(c_{2}=\binom{n}{2}=\frac{1}{2} n(n-1)\). Therefore, the equation

\[
1+n+\frac{1}{2} n(n-1)=79
\]

needs to be solved. After multiplying by 2 and subtracting 79, we obtain the quadratic equation

\[
n^{2}+n+156=0
\]

This can be solved using the \(p q\)-formula, and we get the solutions

\[
n_{1}=12, \quad n_{2}=-13
\]

Clearly, only \(n=12\) is meaningful. This is the only number that satisfies the condition given in a).

b) The monomial \(x^{0}\) arises if there is a number \(k\) such that \(k\) times \(x^{2}\) and \(n-k\) times \(\frac{1}{x}\) are multiplied together, and

\[
2 k+(-1)(n-k)=0
\]

is satisfied. Since no terms are negative, they cannot cancel each other out. Equation (2) is thus the only condition that \(n\) must satisfy. Additionally, it is equivalent to \(n=3 k\). Therefore, \(n\) must be divisible by 3.

c) The number 12 is the only natural number that satisfies both the conditions given in a) and b).