Solve the following equation in the set of natural numbers:

$$
x^{3}-y^{3}=x y+61
$$
Transform the equation as follows:

$$
(x-y)\left(x^{2}+x y+y^{2}\right)=x y+61
$$

Let's introduce the new variables $x-y=a$ and $x y=b$. Since

$$
x^{2}+x y+y^{2} \geq 0
$$

and in the case of equality

$$
0 \neq x y+61
$$

therefore

$$
x^{2}+x y+y^{2}>0 \quad *
$$

Thus, given $x y \geq 0$, $a$ is positive, and $b$ is a non-negative integer.

On the other hand,

$$
x^{2}+x y+y^{2}=(x-y)^{2}+3 x y=a^{2}+3 b
$$

so

$$
a\left(a^{2}+3 b\right)=b+61, a^{3}+3 a b=b+61,3 a b-b=61-a^{3}, b=\frac{61-a^{3}}{3 a-1}
$$

Here, since $a \geq 1$, $3 a-1>0$, thus it is necessary that $61 \geq a^{3}$, which means $a=1,2,3$.

If $a=1$, then $b=30$.

If $a=2$, then $b=\frac{53}{5}$, which is not an integer.

If $a=3$, then $b=\frac{34}{8}$, which is not an integer.

Substituting the only possible $a=1, b=30$ back: $x-y=1$ and $x y=30$. From the first equality, $x=y+1$, substituting this in

$$
y(y+1)=30 y^{2}+y-30=0(y-5)(y+6)=0
$$

Since $y \geq 0$, then $y=5$, so $x=6$. For these values, the desired equality indeed holds, thus the solution is $x=6, y=5$.

Based on the work of Péter Csordás (Kecskemét, Zrínyi I. Secondary School, 8th grade) [^0]

[^0]:    ${ }^{1}$ If zero is not considered a natural number, $x^{2}+x y+y^{2}>0$ is true independently of the condition. This kind of definition changes the $\geq$ type conditions to $>$ in the solution, but this is insignificant from the perspective of the solution.