Find $2$ prime numbers with the given property: The number is one less than a square number. If this cannot be done, prove that it is impossible.
I started working on this problem a few days ago and gave up, but decided to restart today. I have got: $2^2 - 1 = 3$, which is prime, so $2$ is the first prime with this property. I have devised a simple equation that can help:
$x^2 - 1 = p$, where $p$ is a prime number and $x$ is an integer
$x$ must be even as most primes are odd (except $2$, which has already been considered).
$(\text{Anything ending in } 4)^2$ ends in a 6 and $(\text{anything ending in a } 6)^2$ ends in a $6$.
$x^2$ cannot end in a $6$ as then, when $1$ is subtracted, it ends in $5$, which means that number is a multiple of $5$. This does not include $6$ itself as $\sqrt{6}$ is not a square number. So, $x$ can't end in a $4$ or a $6$.
However, I couldn't get past this point. Could anyone offer a solution? Thanks in advance.
Answering this since the question remains unanswered:
As you have already shown, $3$ satisfies this property. In general, if $p$ is a prime with $p=n^2-1$ for some natural $n$, we have that $p=(n+1)(n-1)$ (difference of squares), so either $n+1=1, n-1=1$, or $p$ is not prime (it is the product of two natural numbers).
If $n+1=1$ then $n=0$, so $p=-1$; clearly not a prime.
If $n-1=1$ then $n=2$, so $p=3$.
Since there are no other cases, there is only one prime, $3$, satisfying this property.