TEAM SELECTION EXAMINATION FOR THE 42nd INTERNATIONAL MATH- EMATICAL OLYMPIAD. TURKEY.

Let us consider the equation:
$$5^x = 1 + 4y + y^4$$
We seek integer solutions $(x, y)$.

First, note that $y^4 + 4y + 1$ grows rapidly for large $|y|$, so $5^x$ must also be a perfect power of $5$.

Let us try small integer values for $y$:

For $y = 0$:
$$1 + 4 \cdot 0 + 0^4 = 1$$
So $5^x = 1 \implies x = 0$.
Thus, $(x, y) = (0, 0)$ is a solution.

For $y = 1$:
$$1 + 4 \cdot 1 + 1^4 = 1 + 4 + 1 = 6$$
$6$ is not a power of $5$.

For $y = -1$:
$$1 + 4 \cdot (-1) + (-1)^4 = 1 - 4 + 1 = -2$$
$-2$ is not a power of $5$.

For $y = 2$:
$$1 + 4 \cdot 2 + 2^4 = 1 + 8 + 16 = 25$$
$25 = 5^2$, so $x = 2$.
Thus, $(x, y) = (2, 2)$ is a solution.

For $y = -2$:
$$1 + 4 \cdot (-2) + (-2)^4 = 1 - 8 + 16 = 9$$
$9$ is not a power of $5$.

For $y = 3$:
$$1 + 4 \cdot 3 + 3^4 = 1 + 12 + 81 = 94$$
$94$ is not a power of $5$.

For $y = -3$:
$$1 + 4 \cdot (-3) + (-3)^4 = 1 - 12 + 81 = 70$$
$70$ is not a power of $5$.

For $y = 4$:
$$1 + 4 \cdot 4 + 4^4 = 1 + 16 + 256 = 273$$
$273$ is not a power of $5$.

For $y = -4$:
$$1 + 4 \cdot (-4) + (-4)^4 = 1 - 16 + 256 = 241$$
$241$ is not a power of $5$.

Now, for $y = 5$:
$$1 + 4 \cdot 5 + 5^4 = 1 + 20 + 625 = 646$$
$646$ is not a power of $5$.

For $y = -5$:
$$1 + 4 \cdot (-5) + (-5)^4 = 1 - 20 + 625 = 606$$
$606$ is not a power of $5$.

Now, consider negative $x$:
$5^x$ is always positive, so $1 + 4y + y^4 > 0$.
For $y < 0$, $y^4$ is positive, but $4y$ is negative. For large negative $y$, $y^4$ dominates, so $1 + 4y + y^4$ is positive.

Let us check for $y = -10$:
$$1 + 4 \cdot (-10) + (-10)^4 = 1 - 40 + 10000 = 9961$$
$9961$ is not a power of $5$.

Now, let's try to factor $y^4 + 4y + 1$ to see if it can be a power of $5$ for other integer $y$.

Alternatively, note that for large $|y|$, $y^4$ dominates, so $1 + 4y + y^4 \approx y^4$.
So $5^x \approx y^4$, so $y$ is approximately $\pm 5^{x/4}$, but $x$ must be divisible by $4$ for $y$ to be integer.

Let us try $x = 4$:
$$5^4 = 625$$
So $y^4 + 4y + 1 = 625$
$y^4 + 4y + 1 - 625 = 0$
$y^4 + 4y - 624 = 0$

Try $y = 4$:
$4^4 + 4 \cdot 4 - 624 = 256 + 16 - 624 = 272 - 624 = -352$

Try $y = 5$:
$5^4 + 4 \cdot 5 - 624 = 625 + 20 - 624 = 645 - 624 = 21$

Try $y = 6$:
$6^4 + 4 \cdot 6 - 624 = 1296 + 24 - 624 = 1320 - 624 = 696$

Try $y = -4$:
$(-4)^4 + 4 \cdot (-4) - 624 = 256 - 16 - 624 = 240 - 624 = -384$

Try $y = -5$:
$625 - 20 - 624 = 605 - 624 = -19$

Try $y = -6$:
$1296 - 24 - 624 = 1272 - 624 = 648$

No integer solution for $y$.

Now, for $x = 1$:
$5^1 = 5$
$y^4 + 4y + 1 = 5$
$y^4 + 4y - 4 = 0$
Try $y = 1$: $1 + 4 - 4 = 1$
$y = -1$: $1 - 4 - 4 = -7$
$y = 2$: $16 + 8 - 4 = 20$
$y = -2$: $16 - 8 - 4 = 4$
No integer solution.

For $x = -1$:
$5^{-1} = 1/5$
$y^4 + 4y + 1 = 1/5$
No integer solution.

Thus, the only integer solutions are $(x, y) = (0, 0)$ and $(x, y) = (2, 2)$.

Final Answer:
All integer solutions are $(x, y) = (0, 0)$ and $(x, y) = (2, 2)$.