## Task 31/77

Decompose the number 100 in all possible ways into a sum of four natural numbers such that the following condition is satisfied:

If a natural number is added to the first summand, the same number is subtracted from the second summand, the third summand is multiplied by this number, and the fourth summand is divided by it, then all the results are equal.
Let $a, b, c, d$ be four natural numbers that satisfy the conditions of the problem. Then the following equations hold:

$$
a+b+c+d=100 \quad ; \quad a+x=b-x=c x=\frac{d}{x}
$$

where $x$ is also a natural number and $x \neq 0$. From the second equations, we obtain

$$
2 c x=a+x+b-x=a+b \quad ; \quad d=c x^{2}
$$

Substituting this into the first equation, we get

$$
2 c x+c+c x^{2}=c\left(x^{2}+2 x+1\right)=c(x+1)^{2}=100
$$

Since $100=10^{2}$, $c$ must also be a perfect square. We therefore decompose the number 100 into all possible products of two perfect squares:

$$
100=10^{2} \cdot 1^{2}=5^{2} \cdot 2^{2}
$$

Thus, the numbers $c_{1}=1 ; c_{2}=4 ; c_{3}=25 ; c_{4}=100$ with $x_{1}=9 ; x_{2}=4 ; x_{3}=1, x_{4}=0$ are possible, where $x_{4}=0$ and thus $c_{4}=100$ can be immediately ruled out. For $c_{1}=1, x_{1}=9$, we get $a_{1}=0, b_{1}=18, d_{1}=81$, the remaining values $c_{2}=4, x_{2}=4$ and $c_{3}=25, x_{3}=1$ yield $a_{2}=12, b_{2}=20, d_{2}=64$ and $a_{3}=24, b_{3}=$ $26, d_{3}=25$. Therefore, there are exactly three such decompositions:

$$
0+18+1+81=100 \quad ; \quad 12+20+4+64=100 \quad ; \quad 24+26+25+25=100
$$