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
"Percentage" is a bad word, you mean something like natural density.
For that it's easy, as the limit exists, you just compute
$$\lim_{n\to\infty} {\#\{m^2 < n : m\in\Bbb N\}\over n}$$
The numerator is bounded above by $\sqrt{n}$ so we get
$$0\le\lim_{n\to\infty} {\#\{m^2 < n : m\in\Bbb N\}\over n}\le \lim_{n\to\infty} {1\over\sqrt n}=0$$
Best Answer
If $p_1$ through $p_n$ form the unique primes in the prime factorization of $k$ (so not including multiplicities) we can use the following definition:
$$\varphi(k) = k\left(1 - \frac1{p_1}\right)\left(1 - \frac1{p_2}\right)\cdots\left(1 - \frac1{p_n}\right)$$
And the knowledge that each prime in the factorization of $k$ occurs twice in $k^2$, without introducing any new primes to find out:
$$\varphi(k^2) = k^2\left(1 - \frac1{p_1}\right)\left(1 - \frac1{p_2}\right)\cdots\left(1 - \frac1{p_n}\right)$$
$$\varphi(k^2) = k\varphi(k)$$