Find all primes $p$, such that there exist positive integers $x$, $y$ which satisfy $$\begin{cases} p + 49 = 2x^2\\ p^2 + 49 = 2y^2\\ \end{cases}$$
Problem
Source: 2022 Bulgarian Spring Math Competition, Problem 9.3
Tags: algebra, system of equations, prime
VicKmath7
27.03.2022 14:30
Note that $p>y$, otherwise $y \leq 7$ and case check. Now, subtracting the two equations, we obtain $p(p-1)=2(y-x)(y+x)$. Obviously $p=2$ doesn't work, and $p$ can't divide $y-x$. So $p|y+x$ and if $x+y$ is not equal to $p$, then $x+y \geq 2p > p+y$, so $x>p$ and $p+49>2p^2$, case check. So $x+y=p, y-x=\frac {p-1}{2}$, calculate $x, y$ in terms of $p$ and plug in the system, done.
grupyorum
27.03.2022 20:00
Way older than USAMO 2022. Very similar idea in Bundeswettbewerb Mathematik 1997.
Isolemma
11.05.2022 10:07
Even removing the requirement that $p$ is prime, this is an elliptic curve and it is possible to determine all integral points on it. More precisely, we can rewrite it as $(2x^2 - 49)^2 = 2y^2 - 49$. Multiplying by $2x^2$ gives us $$Y^2 = X^3 - 98 X^2 + 2450 X$$where $X = 2x^2$ and $Y = 2xy$. A computer algebra system can be used to compute the Mordell-Weil rank, which is equal to $2$, and to determine all the integral points (up to sign and ignoring the infinity point): $$(X, Y) = (0, 0), (25, 125), (32, 104), (49, 49), (50, 50), (72, 204), (98, 490), (9800, 965300).$$Translating back into $(x, y, p)$, we find the following solutions:$$(x, y, p) = (0, 35, -49), (4, 13, -17), (5, 5, 1), (6, 17, 23), (7, 35, 49), (70, 6895, 9751).$$Among these, only $(6, 17, 23)$ gives prime value of $p$.