For a quintet of integers, it holds that when we add one to the first, square the second, subtract three from the third, multiply the fourth by four, and divide the fifth by five, we get the same result each time.

Find all such quintets of numbers whose sum is 122.

(L. Dedková)

Hint. What properties does the number have that is the result of the described operations?
The result of the described operations with each of the five numbers is always the same, and we denote this by $n$. Then the first to fifth number is

$$
n-1, \quad \sqrt{n}, \quad n+3, \quad \frac{n}{4}, \quad 5 n .
$$

From the second and fourth numbers, we derive important constraints, namely that $n$ is a square of a natural number and that it is divisible by four.

The smallest number with these properties is $n=4$; in this case, the sum of the numbers would be

$$
3+2+7+1+20=33
$$

which is too small. The next number with the given properties is $n=16$; in this case, the sum of the numbers would be

$$
15+4+19+4+80=122
$$

which is the required result. Further options do not need to be tried, as the total sum increases with increasing $n$. The only solution to the problem is the quintuple $15,4,19,4,80$.

Another solution. Let the quintuple of the sought numbers be $a, b, c, d, e$. From the problem statement, it follows that

$$
a+1=b^{2}=c-3=4 d=\frac{e}{5} .
$$

For $d=\frac{b^{2}}{4}$ to be an integer, $b^{2}$ must be a multiple of 4, so $b$ must be a multiple of 2. We express this as $b=2 k$, for some integer $k$. From the given equations, we deduce that

$$
a=4 k^{2}-1, \quad b=2 k, \quad c=4 k^{2}+3, \quad d=k^{2}, \quad e=20 k^{2}
$$

The sum of all these numbers is $s=29 k^{2}+2 k+2$. For sufficiently large absolute values of $k$ (regardless of the sign), the sum will be greater than 122 and will continue to increase sharply. By trial, we find the following:

| $k$ | $\leqq-3$ | -2 | -1 | 0 | 1 | 2 | $\geqq 3$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $s$ | $\geqq 257$ | 114 | 29 | 2 | 33 | 122 | $\geqq 269$ |

The sum is equal to 122 precisely for $k=2$, which determines the quintuple of numbers

$$
a=15, \quad b=4, \quad c=19, \quad d=4, \quad e=80 .
$$