1. **Problem statement:** Find all prime numbers $p$ such that $p-3$ is a perfect square. 2. **Set up the equation:** Let $p-3 = n^2$ where $n$ is an integer. 3. **Rewrite:** Then $p = n^2 + 3$. 4. **Check for prime $p$:** We want $p = n^2 + 3$ to be prime. 5. **Test values of $n$:** - For $n=0$, $p=0^2+3=3$ which is prime. - For $n=1$, $p=1+3=4$ not prime. - For $n=2$, $p=4+3=7$ prime. - For $n=3$, $p=9+3=12$ not prime. - For $n=4$, $p=16+3=19$ prime. - For $n=5$, $p=25+3=28$ not prime. - For $n=6$, $p=36+3=39$ not prime. - For $n=7$, $p=49+3=52$ not prime. - For $n=8$, $p=64+3=67$ prime. - For $n=9$, $p=81+3=84$ not prime. - For $n=10$, $p=100+3=103$ prime. 6. **Check if more primes exist:** As $n$ grows, $p=n^2+3$ grows quickly and primality becomes less frequent. The problem does not specify an upper bound, so the primes found for small $n$ are: $$p = 3, 7, 19, 67, 103$$ 7. **Conclusion:** The primes $p$ such that $p-3$ is a perfect square are $3, 7, 19, 67,$ and $103$ for $n=0,2,4,8,10$ respectively. **Note:** Larger $n$ values can be tested similarly, but these are the initial primes satisfying the condition.