(a) Find all prime numbers p such that $$4p^2+1$$ and $$6p^2+1$$ are also primes.
(b) Determine real numbers x, y, z, u such that
$$xyz+xy+yz+zx+x+y+z=7$$,
$$yzu+yz+zu+uy+y+z+u=9$$,
$$zux+zu+ux+xz+z+u+x=9$$,
$$uxy+ux+xy+yu+u+x+y=9$$.

For part (a), checking small primes shows $$p=5$$ gives $$4(5)^2+1=101$$ and $$6(5)^2+1=151$$, both prime, while any prime $$p>5$$ must be of the form $$5m+1$$ or $$5m+4$$, and in either case one of $$4p^2+1$$ or $$6p^2+1$$ becomes a multiple of 5 greater than 5, hence not prime. So $$p=5$$ is the only solution. For part (b), each equation factors nicely, e.g. the first as $$(z+1)(x+1)(y+1)=8$$, and similarly $$(u+1)(y+1)(z+1)=10$$, $$(x+1)(z+1)(u+1)=10$$, $$(y+1)(u+1)(x+1)=10$$; multiplying all four and taking a fourth root gives $$(x+1)(y+1)(z+1)(u+1)=20$$, and dividing this by each individual equation gives $$u+1=2$$, $$x+1=2$$, $$y+1=2$$, $$z+1=2$$, so $$x=y=z=1$$ and $$u=2$$.