5. Find all integer solutions of the equation $x^{2}+73=y^{2}$.

Solve each problem independently on the attached paper, and each problem on its own page.

Do not sign the sheets, write only your code.

Write your answers with a pen clearly and neatly. Each problem's solution will be graded from 0 to 6 points.

You have 120 minutes to solve the problems.

3rd Mathematical Competition for Students of Secondary Technical and Vocational Schools, Ljubljana, April 12, 2003

## PROBLEMS FOR 4th GRADE
5. We rewrite the equation as $y^{2}-x^{2}=73$ or $(y+x)(y-x)=73$. The number 73 is a prime number, so it can only be factored in four ways: $73 \cdot 1, 1 \cdot 73, -73 \cdot(-1)$, and $-1 \cdot(-73)$. This leads us to four systems of equations, which we solve:

$$
\begin{array}{rrrr}
y+x=73 & y+x=-73 & y+x=1 & y+x=-1 \\
\frac{y-x=1}{y=37} & \frac{y-x=-1}{y=37} & \frac{y-x=73}{y=-37} & \frac{y-x=-73}{y=-37} \\
x=36 & x=-36 & x=-36 & x=36
\end{array}
$$

![](https://cdn.mathpix.com/cropped/2024_06_07_06886e410b39c0216094g-12.jpg?height=62&width=1642&top_left_y=226&top_left_x=207)

![](https://cdn.mathpix.com/cropped/2024_06_07_06886e410b39c0216094g-12.jpg?height=60&width=1639&top_left_y=278&top_left_x=208)

Conclusion: $x$ and $y$ must satisfy one of the four systems: ....................... 1 point

$$
\begin{array}{llll}
y+x=73 & y+x=-73 & y+x=1 & y+x=-1 \\
y-x=1 & y-x=-1 & y-x=73 & y-x=-73
\end{array}
$$

Solutions to the systems $(36,37),(-36,-37),(-36,37)$, and $(36,-37) \ldots \ldots \ldots . \ldots . \ldots . . . . . . . . . . . . .$.

NOTE: only three pairs are written .......................................................................................................................

only one or two pairs are written .........................................................................................

## Fourth Year