Prime Numbers And Perfect Squares Find all primes $p$ and $q$ such that $p^2 + 7pq + q^2$ is a perfect square.
One obvious solution is $p = q$ and under such a situation all primes $p$ and $q$ will satisfy.
Further if $p \neq q$ then we can assume without the loss of generality that $p > q$. Assuming this and that there exists at least one such perfect square I have tried to show some contradiction modulo $4$ as any odd perfect square leaves a remainder of $1$ when divided by $4$, but it is not working. However I firmly believe that $p = q$ is the only solution, but I have failed to prove this.
 A: Since the equation $p^2+7pq+q^2=n^2$ is symmetric in $p$ and $q$, we may assume without loss of generality that $p\lt q$ (setting aside the obvious solutions with $p=q$).  For reasons that will become clear in a moment, it's easy enough to check for solutions with $p\lt q\le7$ (there are none).
Multiplying both sides of the equation by $4$, it's possible to rewrite it as
$$(2p+7q+2n)(2p+7q-2n)=45q^2$$
Now the prime $q$ cannot divide both factors on the left hand side, since the sum of those two factors is congruent to $4p$ mod $q$.  But $2p+7q-2n\lt9q\lt q^2$, so we must have $2p+7q+2n=aq^2$ and $2p+7q-2n=b$ with $ab=45$.  Multiplying each of these by $a$ and summing (to eliminate $n$), we have
$$4ap+14aq=a^2q^2+45$$
or
$$4ap+4=a^2q^2-14aq+49=(aq-7)^2$$
which solves to
$$q={7+2\sqrt{ap+1}\over a}$$
(We can disregard the negative square root, since $q\gt7$.)  This implies $ap+1$ is a square, which clearly implies $ap+1=k^2$ for some integer $k\ge2$ (since $ap\ge2$).  But now the inequality $q\gt p$ becomes $7+2k\gt k^2-1$, or $(k-1)^2\lt9$.  Thus $k\lt4$, so there are just two possibilities:  $k=2$ or $k=3$, correpsponding to $ap+1=4$ and $ap+1=9$.  But since $p$ is primes and $a$ is a divisor of $45$ (hence odd), the only possibility is $p=3$ and $a=1$, which gives $q=7+2\sqrt4=11$.  And indeed
$$3^2+7\cdot3\cdot11+11^2=361=19^2$$
And that's it:  The only prime pairs $(p,q)$ for which $p^2+7pq+q^2$ is a square are $(3,11)$, $(11,3)$, and pairs of the form $(p,p)$.
Remark (added later, after a restless night's sleep):  When I first wrote up this answer, I did the scratchwork to check that there are no solutions with $p\lt q\le7$.  This did the job, to be sure, but it felt a little unsatisfying.  It only later (in the middle of the night) occurred to me to think mod $8$:  If $p\lt q$, then $q$, and hence $n$, are both odd, so that $q^2\equiv n^2\equiv1$ mod $8$, which reduces the equation to $p(p-q)\equiv0$ mod $8$.  From this it's easy to see that $p\equiv q$ mod $8$, which implies $q\ge2+8=10$.
A: Let$p^2+7\times p\times q+q^2=m^2$
$\implies (p+q)^2+5\times p\times q=m^2$
$\implies 5\times p\times q=(m+p+q)(m-p-q)$.
Clearly $m+p+q>m-p-q$
$\implies m+p+q=1,5,p,q,5\times p,5\times q,p\times q,5\times p\times q$
(1)
& $m-p-q=5 \times p \times q,p \times q,5 \times p,5 \times q,p,q,5,1$     
....(2)
Observe $m+p+q=1,5,p,q$ are not possible as $p,q$ are at least 2 & $m+p+q=p,q\implies m+p=0,m+q=0$
From (1)&(2)
$m=\frac{5 \times p+q}{2},\frac{5 \times p\times q+1}{2}$
.
Consider $m+p+q=5\times p$&$m-p-q=q$
$\implies p=q$
Similarly second simmatric eq leads same result.
Take $m+p+q=p \times q$ & $ m-p-q=5$.
$\implies 2(p+q)=p \times q-5
$\implies (p-2)(q-2)=9$
$\implies p=q=5 or (p,q)=(3,11),(11,3)$
So all solutions are$(p,q),(3,11),(11,3)$
A: Of course $p=q$ works for any $p$. Suppose $p\lt q$. Then because
$$
p^2+7pq+q^2=n^2
$$
we have $3p\lt n\lt3q$.
If $q\mid2p+7q-2n$ and $q\mid2p+7q+2n$, then $q\mid 4p$. Thus, $q$ can only divide one of them.
Since
$$
(2p+7q-2n)(2p+7q+2n)=45q^2
$$
we have either
$$
q^2\mid2p+7q-2n\quad\text{or}\quad q^2\mid2p+7q+2n
$$
Thus, $q^2\le2p+7q-2n\lt9q$ or $q^2\le2p+7q+2n\lt15q$. In either case, $q\lt15$. Checking the $15$ cases where $1\lt p\lt q\lt15$ gives that the only solution is $(3,11)$.
A: COMMENT.- Just for fun, prove easily the impossibility for twin numbers and other couples of primes.
First, not possible for $p=2\lt q$. In fact $$q^2+14q+4=z^2\Rightarrow z^2\ge55\Rightarrow z\ge 8$$ Actually $\color{red}{z\ge 9}$ because $z$ must be odd. The discriminant of $q^2+14q+4-z^2=0$ is $ 45+z^2=t^2$ so making $t=z+h$ one has $45=2zh+h^2\ge18h+h^2$. It is clear how to finish.
$$***$$
 Now with the twin numbers $(p,q)=(p,p+2)$  where $p\ge3$. One has
$$p^2+7p(p+2)+(p+2)^2=9p^2+18p+4=z^2$$ 
Because of $p\ge3$ we have $z\ge 12$ so $\color{red}{z\ge 13}$ because $z$ should be odd.
The equation in $3p$
$$(3p)^2+6(3p)+4-z^2=0$$ has discriminant $$9-4+z^2=t^2\iff5+z^2=t^2$$ This is impossible because $z$ being greater or equal than $13$   the minimal difference  $t^2-z^2$ must be greater or equal than $ 14^2-13^2=27.$
$$***$$
There are much primes of the form $(p,q)=(p,p+4)$, for example $(3,7),(7,11),(13,17)$,…… For these we have $$p^2+7p(p+4)+(p+4)^2=z^2=9p^2+36p+16=z^2\Rightarrow z^2\ge 205\Rightarrow \color{red}{z\ge 15}$$ The equation 
$$(3p)^2+12(3p)+16-z^2=0$$ has discriminant $36-16+z^2=t^2\iff 20=t^2-z^2$ but then we have $20\ge 16^2-15^2=31$, absurde.
Wants someone to try to see the case $(p,q)=(p+2h)\large ?$ This includes all the odd primes. 
A: If $p^2+q^2+7pq = r^2$ ($r$ being any integer), then $(p+q)^2 + 5pq = r^2$. So $5pq = r^2-(p+q)^2 = (r+p+q)(r-p-q)$. 
Since $p, q$ and $5$ are all prime, it follows that one of the factors on the right-hand side is equal to one of them, and the other factor is the product of the other two. As clearly visible, $r+p+q$ is greater than any of the numbers $p, q$ and $5$. So it must be the product of two of those numbers (maybe three too), and the other factor $r-p-q$ must be equal to $p, q$ or $5$ (or $1$). 
Now, different cases arise:
CASE $1$:
If $r-p-q = p$ then $r=2p+q$, and thus original equation becomes $p^2+q^2+7pq = (2p+q)^2$, which simplifies to $p=q$, Same if $r-p-q = q$.
CASE $2$:
If $r-p-q = 5$ then $r = p+q+5$, and the equation $5pq = (r+p+q)(r-p-q)$ becomes $pq = 2(p+q)+5$. You can write this as $(p-2)(q-2)=9$, and the only solutions to this are $p=q=5$ and  $p=3;q=11$ (or the other way round). 
CASE $3$: $r+p+q$ might be equal to the product of all three numbers $5pq$, with $r-p-q = 1$. But then the equation becomes $2p+2q+1 = 5pq$, which is clearly impossible because the right-hand side is visibly greater than the left-hand side (Though there are solutions like $(1,1)$, but $1$ is not prime ).
So, to summarise the whole answer, We can say that only solutions are:
$(p,q)=(p,p),(3,11),(11,3)$
SOURCE
A: Assume $p^2+7pq+q^2=n^2$ with $p\ge q$.
Note that $p^2+2pq+p^2=(p+q)^2$, hence
$$(n+p+q)(n-p-q)=n^2-(p+q)^2=5pq$$
We know the prime factorization of $5pq$ and that $n-p-q<n+p+q$, hence conclude that $n-p-q$ is $\in\{1,5,q,p\text{ (if $5q>p$)},5q\text{ (if $5q<p$)}\}$. Inverstigate these cases one by one.
