Determine all pairs (p,q) of prime numbers for which p2+5pq+4q2 is a perfect square.
Problem
Source: Stars of Mathematics 2023 P1 (junior level)
Tags: number theory, prime numbers, Perfect Squares
26.11.2023 23:45
Completing the square yields (p+2q)2+pq=m2⟹pq=m2−(p+2q)2⟹pq=(m−p−2q)(m+p+2q). WLOG p≥q, then p=m+p+2q⟹m+2q=0. Since q>0⟹m<0. Also, q=m−p−2q⟹3q=m−p. Since p>0 and m<0, then 3q<0 or q<0 which means q is not prime. But q must be prime so there are no solutions.
27.11.2023 00:00
Let p2+5pq+4q2=k2 for k∈N. Then, (p+2q)2+pq=k2⇒(k−p−2q)(k+p+2q)=pq.Now that k+p+2q>max, we must have k-p-2q=1 and k+p+2q=pq. So, k=p+2q+1, yielding 2p+4q+1=pq, that is (p-4)(q-2)=9. From here, we obtain (p,q)=(5,11),(7,5),(13,3) as the only solutions.
27.11.2023 00:09
OHHHHH i forgot the other case im kinda goofy
27.11.2023 12:21
Assume p,q\ne 2. (p+4q)(p+q) is a square so if p\ne q then p+4q=a^2,p+q=b^2\Rightarrow a^2-b^2=3q\Rightarrow (a-b)(a+b)=3q\Rightarrow a-b=1,a+b=3q or a-b=3,a+b=q. If a-b=1,a+b=3q: a=\frac{3q+1}2\Rightarrow p+4q=\left(\frac{3q+1}2\right)^2\Rightarrow 4p+16q=9q^2+6q+1\Rightarrow 4p=9q^2-10q+1=(9q-1)(q-1) which gives solutions (5,11),(7,5),(13,3).