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 $5^x=1⟹x=0$. 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 $(x,y)=(2,2)$.