3. Given arithmetic sequences $\left(x_{i}\right)_{i=1}^{\infty}$ and $\left(y_{i}\right)_{i=1}^{\infty}$ have the same first term and the following property: there exists an index $k(k>1)$, for which the equalities

$$
x_{k}^{2}-y_{k}^{2}=53, \quad x_{k-1}^{2}-y_{k-1}^{2}=78, \quad x_{k+1}^{2}-y_{k+1}^{2}=27 .
$$

hold. Find all such indices $k$.
SOLUTION. Let $c, d$ be the differences of the first and second given arithmetic sequences, respectively. Since according to the problem statement $y_{1}=x_{1}$, the terms of both sequences have the general form

$$
x_{i}=x_{1}+(i-1) c \quad \text { and } \quad y_{i}=x_{1}+(i-1) d
$$

for any index $i$. The difference $x_{i}^{2}-y_{i}^{2}$ can therefore be rewritten as

$$
\begin{aligned}
x_{i}^{2}-y_{i}^{2} & =\left(x_{1}^{2}+2 x_{1}(i-1) c+(i-1)^{2} c^{2}\right)-\left(x_{1}^{2}+2 x_{1}(i-1) d+(i-1)^{2} d^{2}\right)= \\
& =2 x_{1}(i-1)(c-d)+(i-1)^{2}\left(c^{2}-d^{2}\right) .
\end{aligned}
$$

For the index $k$, according to the problem statement, the system of equations holds:

$$
\begin{aligned}
& 53=2 x_{1}(k-1)(c-d)+(k-1)^{2}\left(c^{2}-d^{2}\right), \\
& 78=2 x_{1}(k-2)(c-d)+(k-2)^{2}\left(c^{2}-d^{2}\right), \\
& 27=2 x_{1} k(c-d)+k^{2}\left(c^{2}-d^{2}\right) .
\end{aligned}
$$

We will now add these equations or their multiples in a suitable manner. To eliminate the terms with $x_{1}$, we subtract the sum of equations (2) and (3) from twice equation (1), since the coefficient of $2 x_{1}(c-d)$ then becomes $2(k-1)-(k-2+k)=0$. Since $2 \cdot 53-(78+27)=1$ and $2(k-1)^{2}-(k-2)^{2}-k^{2}=-2$, we obtain the simple equation $1=-2\left(c^{2}-d^{2}\right)$, from which we determine $c^{2}-d^{2}=-\frac{1}{2}$. Substituting this into equations (2) and (3), they transform into the form

$$
\begin{aligned}
& 78=2 x_{1}(k-2)(c-d)-\frac{1}{2}(k-2)^{2}, \\
& 27=2 x_{1} k(c-d)-\frac{1}{2} k^{2} .
\end{aligned}
$$

To eliminate the terms with $x_{1}$ again, we subtract $(k-2)$-times equation $\left(3^{\prime}\right)$ from $k$-times equation $\left(2^{\prime}\right)$; we then solve the resulting equation with the unknown $k$:

$$
\begin{aligned}
78 k-27(k-2) & =-\frac{1}{2}(k-2)^{2} \cdot k+\frac{1}{2} k^{2} \cdot(k-2), \\
51 k+54 & =-\frac{1}{2}\left(k^{3}-4 k^{2}+4 k\right)+\frac{1}{2}\left(k^{3}-2 k^{2}\right), \\
0 & =k^{2}-53 k-54, \\
0 & =(k+1)(k-54) .
\end{aligned}
$$

Since the index $k$ is a natural number, it must be $k=54$. This solves the problem.

Let us add that the problem statement does not require us to investigate whether for the found (unique) value of the index $k$ there exist pairs of sequences satisfying the conditions of the problem. For the sake of interest, let us mention that there are actually infinitely many such pairs of sequences; it is necessary and sufficient that their common first term $x_{1}$ and differences $c, d$ satisfy the conditions $c^{2}-d^{2}=-\frac{1}{2}$ and $x_{1}(c-d)=\frac{55}{4}$. This follows easily from any of the equations (1)-(3) after substituting the values $k=54$ and $c^{2}-d^{2}=-\frac{1}{2}$; you can verify this yourself.

GUIDING AND ADDITIONAL PROBLEMS: