Kvant Math Problem 663

Verified: no
Verdicts: SKIP + SKIP
Solve time: 3m24s
Source on kvant.digital

Problem

Find all prime numbers $p$ for which the number $2^p+p^2$ is also prime.

S. Maizus

Exploration

Consider small prime numbers to understand the behavior of the expression $2^p + p^2$. For $p = 2$, we have $2^2 + 2^2 = 4 + 4 = 8$, which is not prime. For $p = 3$, $2^3 + 3^2 = 8 + 9 = 17$, which is prime. For $p = 5$, $2^5 + 5^2 = 32 + 25 = 57$, which is divisible by $3$ and not prime. For $p = 7$, $2^7 + 7^2 = 128 + 49 = 177$, divisible by $3$ and not prime. For $p = 11$, $2^{11} + 11^2 = 2048 + 121 = 2169$, divisible by $3$ and not prime. This pattern suggests that for odd primes $p > 3$, the sum $2^p + p^2$ may always be divisible by $3$, which would prevent it from being prime. The step most likely to hide an error is confirming this divisibility claim rigorously for all odd primes greater than $3$.

Problem Understanding

The problem asks for all prime numbers $p$ such that $2^p + p^2$ is also prime. This is a Type A problem because we are asked to find all numbers satisfying a given condition, and the solution must classify all possibilities. The core difficulty lies in handling all odd primes $p > 3$ and proving that none of them can make $2^p + p^2$ prime. From the exploration, the intuitive answer appears to be $p = 3$, because smaller primes fail and larger primes seem to be constrained by modular considerations, particularly modulo $3$.

Proof Architecture

Lemma 1: If $p = 2$, then $2^p + p^2$ is not prime. This follows by direct computation: $2^2 + 2^2 = 8$.

Lemma 2: If $p = 3$, then $2^p + p^2$ is prime. Direct computation gives $2^3 + 3^2 = 17$, which is prime.

Lemma 3: If $p$ is an odd prime greater than $3$, then $2^p + p^2$ is divisible by $3$. This relies on the congruences $2^p \equiv 2 \pmod 3$ and $p^2 \equiv 1 \pmod 3$, giving $2^p + p^2 \equiv 0 \pmod 3$. This is the hardest lemma, as a careless application of modular arithmetic could fail for small $p$.

Solution

Lemma 1 is verified by computing $2^2 + 2^2 = 8$, which is not prime.

Lemma 2 is verified by computing $2^3 + 3^2 = 8 + 9 = 17$, which is prime.

For Lemma 3, let $p$ be an odd prime greater than $3$. Then $p$ is congruent to $1$ or $2$ modulo $3$. Compute $2^p$ modulo $3$. Since $2^1 \equiv 2 \pmod 3$ and $2^2 \equiv 1 \pmod 3$, the powers of $2$ modulo $3$ cycle with period $2$. Therefore, if $p \equiv 1 \pmod 2$, then $2^p \equiv 2 \pmod 3$. Next, $p^2 \equiv 1 \pmod 3$ because $p \equiv 1$ or $2 \pmod 3$. Adding these, $2^p + p^2 \equiv 2 + 1 \equiv 0 \pmod 3$. Since $2^p + p^2 > 3$, it cannot be prime.

The only remaining candidate is $p = 3$, which has already been verified.

Hence, the complete list of primes $p$ such that $2^p + p^2$ is prime is

$\boxed{3}.$

Verification of Key Steps

The critical step is the modular argument for odd primes $p > 3$. Verify with $p = 5$: $2^5 = 32 \equiv 2 \pmod 3$, $5^2 = 25 \equiv 1 \pmod 3$, sum $32 + 25 = 57 \equiv 0 \pmod 3$. For $p = 7$: $2^7 = 128 \equiv 2 \pmod 3$, $7^2 = 49 \equiv 1 \pmod 3$, sum $128 + 49 = 177 \equiv 0 \pmod 3$. These examples confirm the lemma's validity. The cycle length of $2$ modulo $3$ is correctly applied.

Alternative Approaches

An alternative approach is to consider parity and small primes separately. Noting that for $p$ even, $p = 2$ gives a composite sum, and for odd $p > 3$, $2^p + p^2$ is always divisible by $3$, immediately excluding all larger primes. This approach is equivalent in rigor but shorter in writing. The chosen method is preferable because it explicitly identifies the modular constraint responsible for all exclusions, making the argument transparent and general.