74(1184). Find all pairs of natural numbers that satisfy the equation: a) $x^{2}-y^{2}=105$; b) $2 x^{2}+5 x y-12 y^{2}=28$.
Solution. a) Let's factorize the right and left sides of the equation. We have $(x-y)(x+y)=1 \cdot 105=3 \cdot 35=5 \cdot 21=7 \cdot 15$. Since for natural values of $x$ and $y$, $x-y < x+y$, the solution of this equation reduces to finding natural solutions of the following four systems of equations:
$\left\{\begin{array}{l}x-y=1, \\ x+y=105 ;\end{array} \quad\left\{\begin{array}{l}x-y=3, \\ x+y=35 ;\end{array} \quad\left\{\begin{array}{l}x-y=5, \\ x+y=21 ;\end{array} \quad\left\{\begin{array}{l}x-y=7, \\ x+y=15 .\end{array}\right.\right.\right.\right.$

Answer: $(53 ; 52),(19 ; 16),(13 ; 8),(11 ; 4)$.

b) Let's factorize the left side of the equation:

I method.

$$
\begin{aligned}
& 2 x^{2}+5 x y-12 y^{2}=2 x^{2}+8 x y-3 x y-12 y^{2}= \\
& =2 x(x+4 y)-3 y(x+4 y)=(x+4 y)(2 x-3 y)
\end{aligned}
$$

II method. Solve the equation $2 x^{2}+5 x y-12 y^{2}=0$ with respect to $x$, considering $y$ as a constant number. We have $x_{1}=-4 y, x_{2}=$ $=\frac{3}{2} y$. Then

$$
2 x^{2}+5 x y-12 y^{2}=2(x+4 y)\left(x-\frac{3}{2} y\right)=(x+4 y)(2 x-3 y)
$$

By factoring the right side of the equation into two natural factors $(28=1 \cdot 28=2 \cdot 14=4 \cdot 7)$ and considering that for natural $x$ and $y$, $x+4 y \geqslant 5$, the solution of this equation reduces to solving the following systems of equations in natural numbers:

$$
\left\{\begin{array} { l } 
{ x + 4 y = 2 8 , } \\
{ 2 x - 3 y = 1 ; }
\end{array} \quad \left\{\begin{array} { l } 
{ x + 4 y = 1 4 , } \\
{ 2 x - 3 y = 2 ; }
\end{array} \quad \left\{\begin{array}{l}
x+4 y=7 \\
2 x-3 y=4
\end{array}\right.\right.\right.
$$

Of these systems of equations, only the first one has a solution in natural numbers: $x=8, y=5$.

Answer: $(8 ; 5)$.

$75(1185)$. Prove that the expression $\left(a^{2}+b^{2}\right)^{3}$ can be represented as the sum of the squares of two integer expressions.

Solution.

$$
\begin{gathered}
\left(a^{2}+b^{2}\right)^{3}=\left(a^{2}+b^{2}\right)^{2}\left(a^{2}+b^{2}\right)= \\
=\left(a^{2}+b^{2}\right)^{2} a^{2}+\left(a^{2}+b^{2}\right)^{2} b^{2}=\left(a^{3}+a b^{2}\right)^{2}+\left(a^{2} b+b^{3}\right)^{2}
\end{gathered}
$$

$76(1186)$. Find all solutions of the equation

$$
x^{2}+5 y^{2}+4 x y+2 y+1=0
$$

Solution.

$$
\begin{gathered}
x^{2}+5 y^{2}+4 x y+2 y+1=\left(x^{2}+4 x y+4 y^{2}\right)+\left(y^{2}+2 y+1\right)= \\
=(x+2 y)^{2}+(y+1)^{2}=0 .
\end{gathered}
$$

Since the square of any number is always non-negative, then (considering the equality of the sum of squares to zero) we have the system:

$$
\left\{\begin{array}{l}
x+2 y=0, \\
y+1=0,
\end{array} \quad \text { from which } \quad x=2, y=-1\right.
$$

Answer: $(2 ;-1)$.

Remark. In case of difficulty, students can be offered auxiliary problems (see 2, p. 15).
$77(1187)$. Find all pairs $(x ; y)$ of integers that are solutions of the equation

$$
\sqrt{x-\frac{1}{5}}+\sqrt{y-\frac{1}{5}}=\sqrt{5}
$$

Solution. From the condition, it follows that $x \geqslant \frac{1}{5}$ and $y \geqslant \frac{1}{5}$, which means that $x$ and $y$ are natural numbers.

Let's bring the expressions under the square roots on the left side of the equation to a common denominator. We will have:

$$
\sqrt{\frac{5 x-1}{5}}+\sqrt{\frac{5 y-1}{5}}=\sqrt{5}
$$

Multiplying both sides of this equation by $\sqrt{5}$, we get an equivalent equation

$$
\sqrt{5 x-1}+\sqrt{5 y-1}=5
$$

Since the left side of this equation is the sum of non-negative terms, each term is no more than 5. From the conditions $\sqrt{5 x-1} \leqslant 5$ and $\sqrt{5 y-1} \leqslant 5$, we find that $x \leqslant 5$ and $y \leqslant 5$.

Thus, $x$ and $y$ are natural numbers, each of which is no more than 5.

By trial, we find two pairs of natural numbers $(1 ; 2)$ and $(2 ; 1)$, which satisfy the given equation.

Answer: $(1 ; 2),(2 ; 1)$.