IMO Team Selection Contest

Answer: $(2^0+1,2^1+1,2^2+1,2^3+1,2^4+1)$.

Clearly $(2^0+1,2^1+1,2^2+1,2^3+1,2^4+1)$ is a perfect tuple with length 5. Show in the rest that there are no other perfect tuples with length 5 or larger.

For that, let $(a^m+1,a^{m+1}+1,\dots ,a^n+1)$ be an arbitrary perfect tuple with length at least 5. There must exist at least two odd exponents among $m,m+1,\dots ,n$; let $k$ and $k+2$ be the two largest odd exponents. As $a^k+1$ and $a^{k+2}+1$ are prime powers while having a common divisor $a+1$, these two integers must be powers of the same prime $q$. Thus the larger of them, $a^{k+2}+1$, is divisible by the smaller one, $a^k+1$, which shows that $a^k+1$ divides also the difference $a^2\cdot (a^k+1)-(a^{k+2}+1)=a^2-1$. Hence $a^k+1\le a^2-1$, implying $k<2$. So $k=1$ as $k$ is odd. By choice of $k$, the only odd exponents in our perfect tuple are 1 and 3 and the tuple is of the form $(a^0+1,a^1+1,a^2+1,a^3+1,a^4+1)$.

As $a+1$ and $a^3+1$ are powers of the same prime number $q$, also the ratio $\frac{a^3+1}{a+1}=a^2-a+1$ is a power of $q$. Note that $a^2-a+1\ge 2a-a+1=a+1$ by $a\ge 2$, hence $a^2-a+1$ is divisible by $a+1$. Thus the difference $(a^2-a+1)-(a+1)(a-2)=3$ is divisible by $a+1$. This filters out the only possibility $a=2$.