Argentina_2018

Clearly $a$ must be a prime (set $x=0$). More exactly $a$ is an odd prime because $a=2$ is not a solution. Furthermore $a+1$ must be a power of 2. Indeed suppose that $a+1$ has an odd prime divisor $p$. Note that $p\le \frac{a+1}{2}$ as $a+1$ is even. Set $x=\frac{1}{2}(p-1)$; it is clear that $0<x<a$. We have

$$4x^2+a=4\cdot \frac{1}{4}(p-1{)}^2+a=p(p-2)+(a+1).\text{ Since }p\text{ divides }a+1,\text{ it also divides }4x^2+a.$$ addition, $p<a<4x^2+a$, hence $4x^2+a$ is composite.

The primes $3=2^2-1$ and $7=2^3-1$ satisfy the conditions. The values of $4x^2+3$ for $x=0,1,2$ are the primes $3,7,19$; the values of $4x^2+7$ for $x=0,1,2,3,4,5,6$ are the primes $7,11,23,43,71,107,151$. We show that $a=3$ and $a=7$ are the only solutions by rejecting all $a=2^m-1$ with $m\ge 4$. For numbers of this form consider $a+9=(a+1)+8$. This even number is not a power of 2 (powers of 2 greater than 8 cannot differ by 8). Let $q\le \frac{a+9}{2}$ be an odd prime divisor of $a+9$. Set $x=\frac{1}{2}(q-3)$: $$\text{note that }x\text{ may be zero but is less than }a.\text{ Now }4x^2+a=4\cdot \frac{1}{4}(q-3{)}^2+a=q(q-6)+(a+9),\text{ so }q$$ divides $4x^2+a$. Also $q\le \frac{a+9}{2}<a$ as $a\ge 15$, hence $q<a\le 4x^2+a$. Then $4x^2+a$ is composite, which completes the proof. The answer is $a=3$ and $a=7$.