Then $a^2 + b^2 \equiv 2 \pmod{4}$, so $14\varphi^2(a) - \varphi(ab) + 22\varphi^2(b) \equiv 2 \pmod{4}$. If $ab$ has at least two distinct prime factors and $a, b \ge 3$, then $\varphi(ab) \equiv 0 \pmod{4}$, implying $0 \equiv 2 \pmod{4}$, contradiction. 
 Subcase 1.1: Both a and b are equal to 1.Then $35 = 2$, contradiction. Subcase 1.2: Exactly one of a or b is equal to 1.WLOG $b = 1$. Then $14\varphi^2(a) - \varphi(a) + 22 = a^2 + 1$. If $8 \mid \varphi(a)$, then $22 \equiv a^2 + 1 \pmod{8} \iff a^2 \equiv 5 \pmod{8}$, which is impossible. So $\nu_2 (\varphi(a)) \le 2$, hence $a$ must have at most $2$ odd prime factors. If $a$ has $2$ odd prime factors $p$ and $q$, then $14\varphi^2(a) = 14\left(\left(\frac{p-1}{p}\right)\left(\frac{q-1}{q}\right)\right)^2 a^2 \ge 14\left(\frac{2}{3} \cdot \frac{4}{5}\right)^2 a^2 = \frac{896}{225}a^2 > 3a^2 > a^2 + a$, and $14\varphi^2(a) - \varphi(a) + 22 > a^2 + a - a + 22 = a^2 + 22 > a^2 + 1$, contradiction. If $a$ has $1$ odd prime factor $p$, then $14\varphi^2(a) = 14\left(\frac{p-1}{p}\right)^2 a^2 \ge 14\left(\frac{2}{3} \right)^2 a^2 = \frac{56}{9}a^2 > 3a^2 > a^2 + a$, and $14\varphi^2(a) - \varphi(a) + 22 > a^2 + a - a + 22 = a^2 + 22 > a^2 + 1$, contradiction. Since $22 \equiv 14 \pmod{8}$ and $22 > 14$, the same argument can be made for $a = 1$ to show that there are no solutions. Subcase 1.3: a and b are both not equal to 1.Then $a, b \ge 3$, hence, $ab$ must be the power of an odd prime. Since we assumed that $\nu_p (gcd(a, b)) \le 1$ for all primes $p$, then either $(a, b) = (p^k, p)$ or $(p, p^k)$ for some positive integer $k$ and odd prime $p$. Note that in both cases, $p-1 \mid \varphi(a), \varphi(b), \varphi(ab)$, so $p-1 \mid a^2 + b^2 = p^{2k} + p^2 = p^2(p^{2k-2}+1) \iff p-1 \mid p^{2k-2}+1 \iff p-1 \mid 2 \iff p = 3$. We see that $14\varphi^2(3) - \varphi(3^{k+1}) + 22\varphi^2(3^k) > 14\varphi^2(3^k) - \varphi(3^{k+1}) + 22\varphi^2(3) = 56 \cdot 3^{2k-2} - 2 \cdot 3^k + 88 > 6 \cdot 3^{2k} - 2 \cdot 3^k + 88 = 4 \cdot 3^{2k} + (2 \cdot 3^{2k} - 2 \cdot 3^k) + 88 > 4 \cdot 3^{2k} + 88 > 3^{2k} + 9 = a^2 + b^2$. Thus, there are no solutions to either case. Thus, there are no solutions when both $a$ and $b$ are odd.

Let $a = 2c$ and $b = 2d$ where $c, d$ are positive integers. If $d = 1$, then $b = 2$ and $14\varphi^2(2c) - \varphi(4c) + 22 = 4c^2 + 4$. If $c = 1$, then $14 - 2 + 22 = 8 \iff 34 = 8$, contradiction. If $c \ge 2$, then $4 \mid \varphi(4c), \varphi^2(2c)$, so $2 \equiv 0 \pmod{4}$, contradiction. The same argument can be made to show there are also no solutions for $c = 1$. Thus, $c, d > 1$, and $a, b > 2$. Since we assumed $\nu_p (gcd(a, b)) \le 1$ for all primes $p$, at least one of $c$ or $d$ must be odd. Subcase 2.1: c and d have different parities.Then $c^2 + d^2$ is odd, and $\nu_2(a^2 + b^2) = \nu_2 (4(c^2 + d^2)) = 2$. Since $2 \mid \varphi(a), \varphi(b)$, then $\nu_2 (14\varphi^2(a)+22\varphi^2(b)) \ge 3$. Thus, $\nu_2 (\varphi(ab)) = 2$. But one of $c$ or $d$ must be even, so $8 \mid ab$. Let $ab = 2^l \cdot m$, where $l \ge 3$ and $m$ is odd. Then $\varphi(ab) = 2^{l-1} \cdot \varphi(m) \implies \nu_2(\varphi(ab)) = l-1 + \nu_2 (\varphi(m))$ which can only be equal to $2$ if $l = 3$ and $\varphi(m)$ is odd $\iff m = 1$. So $ab = 8$. However, since $a, b > 2$, $ab \ge 9$, contradiction. Thus, there are no solutions for this subcase. Subcase 2.2: c and d have the same parity.Then both $c$ and $d$ must be odd. So $$14\varphi^2(a) - \varphi(ab) + 22\varphi^2(b) = a^2 + b^2$$becomes \begin{align} \nonumber 14\varphi^2(2c) - \varphi(4cd) + 22\varphi^2(2d) = 4(c^2 + d^2)\\ \nonumber 14\varphi^2(c) - 2\varphi(cd) + 22\varphi^2(d) = 4(c^2 + d^2)\\ 7\varphi^2(c) - \varphi(cd) + 11\varphi^2(d) = 2(c^2 + d^2) \end{align}where $c$ and $d$ are odd numbers such that $c, d \ge 3$. If $\nu_2 (\varphi(c))$ and $\nu_2 (\varphi(d))$ are both equal to $1$, then $c$ and $d$ can only have $1$ odd prime factor. So $7\varphi^2(c) \ge 7\left(\frac{2}{3}\right)^2c^2 = \frac{28}{9}c^2 > \frac{5}{2}c^2$ and $11\varphi^2(d) \ge 11\left(\frac{2}{3}\right)^2d^2 = \frac{44}{9}d^2 > \frac{5}{2}d^2$. Then $7\varphi^2(c) - \varphi(cd) + 11\varphi^2(d) > \frac{5}{2}(c^2 + d^2) - cd = 2(c^2 + d^2) + \frac{1}{2}(c-d)^2 \ge 2(c^2 + d^2)$, which is in contradiction with $(1)$. Hence, at least one of $c$ or $d$ must have at least $2$ distinct prime factors, and $cd$ must have at least $2$ distinct prime factors. This also implies $\nu_2(\varphi(cd)) \ge 2$. Thus, at least one of $\varphi(c)$ or $\varphi(d)$ is divisible by $4$. WLOG $4 \mid \varphi(c)$. If $\nu_2 (\varphi(cd)) \ge 4$, then taking modulo $16$ on both sides of $(1)$, we have $11\varphi^2(d) \equiv 2(c^2 + d^2)$. Since $c^2, d^2 \equiv 1, 9 \pmod{16}$, $c^2 + d^2 \equiv 2, 10 \pmod{16}$ and $2(c^2 + d^2) \equiv 4 \pmod{16}$. Hence, $11\varphi^2(d) \equiv 4 \pmod{16} \iff \varphi^2(d) \equiv 12 \pmod{16}$. If $4 \mid \varphi(d)$, then $\varphi^2(d) \equiv 0 \pmod{16}$, contradiction. Thus, $\nu_2(\varphi(d)) = 1$. But this would mean $\varphi^2(d) \equiv 4 \pmod{16}$, contradiction. Hence, $\nu_2(\varphi(cd)) \le 3$. We can apply the same argument to $4 \mid \varphi(d)$ to get $\nu_2(\varphi(cd)) \le 3$, because $7\varphi^2(c) \equiv 4 \pmod{16} \iff \varphi^2(c) \equiv 12 \pmod{16}$. Thus, $\nu_2(\varphi(cd))$ can only be either $2$ or $3$. This implies that $cd$ has at most $3$ distinct prime factors. We have shown earlier that $cd$ must have at least $2$ distinct prime factors, so $cd$ can only have either $2$ or $3$ distinct prime factors. Subsubcase 2.2.1: cd has exactly 2 distinct prime factors.Let those prime factors be $p$ and $q$ such that $p < q$. If $p \ge 5$, then $7\varphi^2(c) \ge 7\left(\frac{4}{5} \cdot \frac{6}{7}\right)^2 c^2 = \frac{576}{125}c^2 > \frac{5}{2}c^2$ and $11\varphi^2(d) \ge 11\left(\frac{4}{5} \cdot \frac{6}{7}\right)^2 d^2 = \frac{6336}{1225}d^2 > \frac{5}{2}d^2$. Then $7\varphi^2(c) - \varphi(cd) + 11\varphi^2(d) > \frac{5}{2}(c^2 + d^2) - cd = 2(c^2 + d^2) + \frac{1}{2}(c-d)^2 \ge 2(c^2 + d^2)$, which is in contradiction with $(1)$. Thus, $p = 3$. If $q \ge 11$, then $7\varphi^2(c) \ge 7\left(\frac{2}{3} \cdot \frac{10}{11}\right)^2 c^2 = \frac{2800}{1089}c^2 > \frac{5}{2}c^2$ and $11\varphi^2(d) \ge 11\left(\frac{2}{3} \cdot \frac{10}{11}\right)^2 d^2 = \frac{400}{99}d^2 > \frac{5}{2}d^2$. Then $7\varphi^2(c) - \varphi(cd) + 11\varphi^2(d) > \frac{5}{2}(c^2 + d^2) - cd = 2(c^2 + d^2) + \frac{1}{2}(c-d)^2 \ge 2(c^2 + d^2)$, which is in contradiction with $(1)$. Hence, $q$ can only be either $5$ or $7$. Note that since at least one of $c$ or $d$ must have at least $2$ distinct prime factors, at least one of $c$ or $d$ must be divisible by $pq$. Subsubsubcase 2.2.1.1: $q = 5$. If $15 \mid c, d$, then if $25 \mid c$ or $25 \mid d$, WLOG $25 \mid c$, then $25 \mid \varphi^2(c), \varphi(cd)$. Since $25 \mid 2(c^2 + d^2)$, we have $25 \mid 11\varphi^2(d) \iff 25 \mid \varphi^2(d) \iff 5 \mid \varphi(d)$. Since $5 \nmid 3 - 1$, this implies that $25 \mid d$. However, we assumed that $\nu_p (gcd(a, b)) \le 1$ for all primes $p$, contradiction. The same argument can be made for $25 \mid d$. Thus, if $15 \mid c, d$, then $\nu_5(c) = \nu_5(d) = 1$, and $c = 3^x \cdot 5, d = 3^y \cdot 5$ for some positive integers $x, y$. Plugging this into $(1)$, we get $$56 \cdot 3^{x-1} - 40 \cdot 3^{x+y-1} + 88 \cdot 3^{y-1} = 50(9^x + 9^y)$$We see that $8 \mid 50(9^x + 9^y) \iff 4 \mid 9^x + 9^y$, which is impossible since $9^x + 9^y \equiv 1 + 1 \equiv 2 \pmod{4}$. So there are no solutions when $15 \mid c, d$. If $15 \mid d$ and $c$ is a power of either $3$ or $5$, then $7\varphi^2(c) \ge 7\left(\frac{2}{3}\right)^2c^2 = \frac{28}{9}c^2 > \frac{5}{2}c^2$ and $11\varphi^2(d) = 11\left(\frac{2}{3} \cdot \frac{4}{5}\right)^2 d^2 = \frac{704}{225}d^2 > \frac{5}{2}d^2$. Then $7\varphi^2(c) - \varphi(cd) + 11\varphi^2(d) > \frac{5}{2}(c^2 + d^2) - cd = 2(c^2 + d^2) + \frac{1}{2}(c-d)^2 \ge 2(c^2 + d^2)$, which is in contradiction with $(1)$. Thus, $15 \mid c$ and $d$ is a power of either $3$ or $5$. If $d = 3$ and $c = 3^x \cdot 5^y$ for some positive integers $x, y$, then plugging this into $(1)$, we get $$448 \cdot 3^{2x-2} \cdot 5^{2y-2} - 8 \cdot 3^x \cdot 5^{y-1} + 44 = 2 \cdot 3^{2x} \cdot 5^{2y} + 18$$$$2 \cdot 3^{2x-2} \cdot 5^{2y-2} + 8 \cdot 3^x \cdot 5^{y-1} = 26$$$$3^{2x-2} \cdot 5^{2y-2} + 4 \cdot 3^x \cdot 5^{y-1} = 13$$If $x \ge 2$, then $3 \mid 13$, contradiction. If $y \ge 2$, then $5 \mid 13$, contradiction. Thus, $x = y = 1$, which gives us $(c, d) = (15, 3)$. It can be easily confirmed that this is a solution to $(1)$. If $d = 3^x$ for some positive integer $x \ge 2$, then $c = 3 \cdot 5^y$. Plugging this into $(1)$, we get $$448 \cdot 5^{2y-2} - 8 \cdot 3^x \cdot 5^{y-1} + 44 \cdot 3^{2x-2} = 18 \cdot 5^{2y} + 2 \cdot 3^{2x}$$
Taking modulo $9$ on both sides, we get $448 \cdot 5^{2y-2} \equiv 0 \pmod{9}$, contradiction. If $d = 5$ and $c = 3^x \cdot 5^y$ for some positive integers $x, y$, then plugging this into $(1)$, we get $$448 \cdot 3^{2x-2} \cdot 5^{2y-2} - 8 \cdot 3^{x-1} \cdot 5^y + 176 = 2 \cdot 3^{2x} \cdot 5^{2y} + 50$$Since the LHS is divisible by $8$, this implies that $4 \mid 3^{2x} \cdot 5^{2y} + 25 \iff 4 \mid 26$, contradiction. If $d = 5^x$ for some positive integer $x \ge 2$, then $c = 3^y \cdot 5$. Plugging this into $(1)$, we get $$448 \cdot 3^{2y-2} - 8 \cdot 5^x \cdot 3^{y-1} + 176 \cdot 5^{2x-2} = 50 \cdot 3^{2y} + 2 \cdot 5^{2x}$$Taking modulo $25$ on both sides, we get $448 \cdot 3^{2y-2} \equiv 0 \pmod{25}$, contradiction. Thus, the only solution when $q = 5$ is $(c, d) = (15, 3)$. Subsubsubcase 2.2.1.2: $q = 7$. If $21 \mid c, d$, then if $49 \mid c$ or $49 \mid d$, WLOG $49 \mid c$, then $49 \mid \varphi^2(c), \varphi(cd)$. Since $49 \mid 2(c^2 + d^2)$, we have $49 \mid 11\varphi^2(d) \iff 49 \mid \varphi^2(d) \iff 7 \mid \varphi(d)$. Since $7 \nmid 3 - 1$, this implies that $49 \mid d$. However, we assumed that $\nu_p (gcd(a, b)) \le 1$ for all primes $p$, contradiction. The same argument can be made for $49 \mid d$. Thus, if $21 \mid c, d$, then $\nu_7(c) = \nu_7(d) = 1$, and $c = 3^x \cdot 7, d = 3^y \cdot 7$ for some positive integers $x, y$. Plugging these into $(1)$ and taking modulo $7$, we get $11 \varphi^2(d) = 176 \cdot 3^{2y} \equiv 0 \pmod{7}$, which is impossible. So there are no solutions when $21 \mid c, d$. If $21 \mid d$ and $c$ is a power of either $3$ or $7$, then $7\varphi^2(c) \ge 7\left(\frac{2}{3}\right)^2c^2 = \frac{28}{9}c^2 > \frac{5}{2}c^2$ and $11\varphi^2(d) = 11\left(\frac{2}{3} \cdot \frac{6}{7}\right)^2 d^2 = \frac{176}{49}d^2 > \frac{5}{2}d^2$. Then $7\varphi^2(c) - \varphi(cd) + 11\varphi^2(d) > \frac{5}{2}(c^2 + d^2) - cd = 2(c^2 + d^2) + \frac{1}{2}(c-d)^2 \ge 2(c^2 + d^2)$, which is in contradiction with $(1)$. Thus, $21 \mid c$ and $d$ is a power of either $3$ or $7$. If $d$ is a power of $7$, then $7\varphi^2(c) = 7\left(\frac{2}{3} \cdot \frac{6}{7}\right)^2 c^2 = \frac{16}{7}c^2 > \frac{13}{6}c^2$ and $11\varphi^2(d) = 11\left(\frac{6}{7}\right)^2 d^2 = \frac{396}{49}d^2 > 8d^2$. Then $7\varphi^2(c) - \varphi(cd) + 11\varphi^2(d) > \frac{13}{6}c^2 + 8d^2 - cd = 2(c^2 + d^2) + \frac{1}{6}c^2 + 6d^2 - cd = 2(c^2 + d^2) + \left(\frac{\sqrt{6}}{6}c - \sqrt{6}d\right)^2 + cd > 2(c^2 + d^2)$, which is in contradiction with $(1)$. So $21 \mid c$ and $d$ must be a power of $3$. Then $3 \mid \varphi(c), \varphi(cd), c^2 + d^2$, so $3 \mid 11\varphi^2(d) \iff 3 \mid \varphi(d)$. Note that $3 \nmid \varphi(3) = 2$, so $9 \mid d$ and $\nu_3(c) = 1$. Hence, $c = 3 \cdot 7^y, d = 3^x$ for some positive integers $x \ge 2$ and $y$. Plugging these into $(1)$, we get $$1008 \cdot 7^{2y-2} - 4 \cdot 3^{x+1} \cdot 7^{y-1} + 44 \cdot 3^{2x-2} = 18 \cdot 7^{2y} + 2 \cdot 3^{2x}$$$$126 \cdot 7^{2y-2} + 26 \cdot 3^{2x-2} = 4 \cdot 3^{x+1} \cdot 7^{y-1}$$If $y \ge 2$, then taking both sides modulo $7$, we see that $26 \cdot 3^{2x-2} \equiv 0 \pmod{7}$, which is impossible. Thus, $y = 1$. Then $126 + 26 \cdot 3^{2x-2} = 4 \cdot 3^{x+1} \iff 63 + 13 \cdot 3^{2x-2} = 2 \cdot 3^{x+1}$. If $x \ge 3$, then $2x - 2 \ge 3$ and $x + 1 \ge 3$, so taking both sides modulo $27$, we see that $63 \equiv 0 \pmod{27}$, contradiction. Hence, $x = 2$. But this gives $63 + 117 = 54$, contradiction. Thus, there are no solutions when $q = 7$. Hence, the only solution to this subsubcase is $(c, d) = (15, 3)$. Subsubcase 2.2.2: cd has exactly 3 distinct prime factors.Since $\nu_2(\varphi(cd)) \le 3$, all $3$ prime factors of $cd$ must be $3 \pmod{4}$, and $\nu_2(\varphi(cd)) = 3$. Note that $2(c^2 + d^2) \equiv 4 \pmod{8}$, so $7\varphi^2(c) - \varphi(cd) + 11\varphi^2(d) \equiv 4 \pmod{8} \iff 7\varphi^2(c) + 11\varphi^2(d) \equiv 4 \pmod{8}$. If $c$ and $d$ both have at least $2$ prime factors, then $8 \mid \varphi^2(c), \varphi^2(d)$ and $7\varphi^2(c) + 11\varphi^2(d) \equiv 0 \pmod{8}$, contradiction. If both $c$ and $d$ have only $1$ prime factor, then $cd$ can only have at most $2$ prime factors, contradiction. Thus, one of $c$ or $d$ must have $1$ prime factor (i.e. be a power of a prime) and the other must have $2$ or $3$ prime factors. If $c$ has only $1$ prime factor, then $7\varphi^2(c) \ge 7\left(\frac{2}{3}\right)^2 c^2 = \frac{28}{9}c^2 > \frac{5}{2}c^2$ and $11\varphi^2(d) \ge 11\left(\frac{2}{3} \cdot \frac{6}{7} \cdot \frac{10}{11}\right)^2 d^2 = \frac{1600}{539}d^2 > \frac{5}{2}d^2$. Then $7\varphi^2(c) - \varphi(cd) + 11\varphi^2(d) > \frac{5}{2}(c^2 + d^2) - cd = 2(c^2 + d^2) + \frac{1}{2}(c-d)^2 \ge 2(c^2 + d^2)$, which is in contradiction with $(1)$. Hence, $d$ must have only $1$ prime factor, let this be $p$. Let the other prime factors of $cd$ be $q$ and $r$. Thus, $qr \mid c$. If $3 \nmid c$, then $7\varphi^2(c) \ge 7\left(\frac{6}{7} \cdot \frac{10}{11} \cdot \frac{18}{19}\right)^2 c^2 = \frac{1166400}{305767}c^2 > \frac{5}{2}c^2$ and $11\varphi^2(d) \ge 11\left(\frac{2}{3}\right)^2 d^2 = \frac{44}{9}d^2 > \frac{5}{2}d^2$. Then $7\varphi^2(c) - \varphi(cd) + 11\varphi^2(d) > \frac{5}{2}(c^2 + d^2) - cd = 2(c^2 + d^2) + \frac{1}{2}(c-d)^2 \ge 2(c^2 + d^2)$, which is in contradiction with $(1)$. Thus, $3 \mid c$, which also implies that one of $p, q, r$ must be $3$. If $c$ and $d$ are relatively prime, then $p \nmid c$ and $p \ne 3$. So $p \ge 7$ and $c$ has only two prime factors, $q$ and $r$. Then $7\varphi^2(c) \ge 7\left(\frac{2}{3} \cdot \frac{6}{7}\right)^2 c^2 = \frac{16}{7}c^2 > \frac{9}{4}c^2$ and $11\varphi^2(d) \ge 11\left(\frac{6}{7}\right)^2 d^2 = \frac{396}{49}d^2 > 6d^2$. So $7\varphi^2(c) - \varphi(cd) + 11\varphi^2(d) > \frac{9}{4}c^2 + 6d^2 - cd = 2(c^2 + d^2) + \frac{1}{4}c^2 + 4d^2 - cd = 2(c^2 + d^2) + \left(\frac{1}{2}c - 2d\right)^2 + cd > 2(c^2 + d^2)$, which is in contradiction with $(1)$. Hence, $c$ and $d$ are not relatively prime. So $pqr \mid c$. If $p^2 \mid c$, then since we assumed that $\nu_p (gcd(a, b)) \le 1$ for all primes $p$, we have $d = p$. Note that $p^2 \mid 7\varphi^2(c), \varphi(cd), 2(c^2 + d^2)$, so $p^2 \mid 11\varphi^2(d) = 11(p - 1)^2 \iff p^2 \mid 11$, which is impossible. Thus, $\nu_p(c) = 1$. Subsubsubcase 2.2.2.1: $d = p$. Taking modulo $p$ on both sides of $(1)$, we get $7\varphi^2(c) + 11 \equiv 0 \pmod{p}$. If $p = 3$, then $\varphi^2(c) \equiv 1 \pmod{3} \iff 3 \nmid \varphi(c) \iff 3 \nmid (q-1)(r-1) \iff q, r \equiv 2 \pmod{3}$. Hence, $q, r \equiv 11 \pmod{12}$. Then $7\varphi^2(c) \ge 7\left(\frac{2}{3} \cdot \frac{10}{11} \cdot \frac{22}{23}\right)^2 c^2 = \frac{11200}{4761}c^2 > \frac{9}{4}c^2$ and $11\varphi^2(d) = 44$. So $7\varphi^2(c) - \varphi(cd) + 11\varphi^2(d) > \frac{9}{4}c^2 + 44 - 3c = 2c^2 + 18 + \left(\frac{1}{2}c - 3\right)^2 + 17 > 2c^2 + 18 = 2(c^2 + d^2)$, which is in contradiction with $(1)$. If $p = 7$, then $11 \equiv 0 \pmod{7}$, contradiction. If $p = 11$, then $11 \mid \varphi(c) \iff 11 \mid (q-1)(r-1)$ which means one of $q-1$ or $r-1$ is divisible by $11$. Since the smallest positive integer that is both $1 \pmod{11}$ and $3 \pmod{4}$ is $23$, at least one of $q$ or $r$ must be greater than or equal to $23$. Then $7\varphi^2(c) \ge 7\left(\frac{10}{11} \cdot \frac{2}{3} \cdot \frac{22}{23}\right)^2 c^2 = \frac{11200}{4761}c^2 > \frac{9}{4}c^2$ and $11\varphi^2(d) = 1100 > 6 \cdot 11^2 = 6d^2$. So $7\varphi^2(c) - \varphi(cd) + 11\varphi^2(d) > \frac{9}{4}c^2 + 6d^2 - cd = 2(c^2 + d^2) + \frac{1}{4}c^2 + 4d^2 - cd = 2(c^2 + d^2) + \left(\frac{1}{2}c - 2d\right)^2 + cd > 2(c^2 + d^2)$, which is in contradiction with $(1)$. If $p = 19$, then $7\varphi^2(c) + 11 \equiv 0 \pmod{19} \iff \varphi^2(c) \equiv 12 \pmod{19}$. But $\left(\frac{12}{19}\right) = \left(\frac{3}{19}\right) = -\left(\frac{19}{3}\right) = -\left(\frac{1}{3}\right) = -1$, so this is impossible. If $p = 23$, then $7\varphi^2(c) + 11 \equiv 0 \pmod{23} \iff \varphi^2(c) \equiv 5 \pmod{23}$. But $\left(\frac{5}{23}\right) = \left(\frac{23}{5}\right) = \left(\frac{3}{5}\right) = -1$, so this is impossible. If $p \ge 31$, then $7\varphi^2(c) \ge 7\left(\frac{30}{31} \cdot \frac{2}{3} \cdot \frac{6}{7}\right)^2 c^2 = \frac{14400}{6727}c^2 > \frac{17}{8}c^2$ and $11\varphi^2(d) \ge 11\left(\frac{30}{31}\right)^2 d^2 = \frac{9900}{961}d^2 > 10d^2$. So $7\varphi^2(c) - \varphi(cd) + 11\varphi^2(d) \ge \frac{17}{8}c^2 + 10d^2 - cd = 2(c^2 + d^2) + \left(\frac{\sqrt{2}}{4}c - 2\sqrt{2}d\right)^2 + cd > 2(c^2 + d^2)$, which is in contradiction with $(1)$. Hence, there are no solutions when $d = p$. Subsubsubcase 2.2.2.2: $d = p^x$ where $x \ge 2$. Then taking both sides of $(1)$ modulo $p$, we get $7\varphi^2(c) \equiv 0 \pmod{p}$. This implies that either $p = 7$, or one of $q - 1$ or $r - 1$ (WLOG $q - 1$) is divisible by $p$. If $p \ge 11$, then $q \ge 23$. So $7\varphi^2(c) \ge 7\left(\frac{10}{11} \cdot \frac{22}{23} \cdot \frac{2}{3}\right)^2 c^2 = \frac{11200}{4761}c^2 > \frac{9}{4}c^2$ and $11\varphi^2(d) = 11\left(\frac{10}{11}\right)^2 d^2 = \frac{100}{11}d^2 > 6d^2$. So $7\varphi^2(c) - \varphi(cd) + 11\varphi^2(d) > \frac{9}{4}c^2 + 6d^2 - cd = 2(c^2 + d^2) + \frac{1}{4}c^2 + 4d^2 - cd = 2(c^2 + d^2) + \left(\frac{1}{2}c - 2d\right)^2 + cd > 2(c^2 + d^2)$, which is in contradiction with $(1)$. Hence, $p = 3$ or $p = 7$. If $p = 7$, then taking both sides of $(1)$ modulo $49$, we have $7\varphi^2(c) \equiv 0 \pmod{49} \iff 7 \mid \varphi(c) \iff 7 \mid (q-1)(r-1)$. So one of $q$ or $r$ must be both $1 \pmod{7}$ and $3 \pmod{4}$, which means that one of $q$ or $r$ is $15 \pmod{28}$. Thus, one of $q$ or $r$ must be at least $43$. So $7\varphi^2(c) \ge 7\left(\frac{6}{7} \cdot \frac{42}{43} \cdot \frac{2}{3}\right)^2 c^2 = \frac{4032}{1849}c^2 > \frac{13}{6}c^2$ and $11\varphi^2(d) = 11\left(\frac{6}{7}\right)^2 d^2= \frac{396}{49}d^2 > 8d^2$. Then $7\varphi^2(c) - \varphi(cd) + 11\varphi^2(d) > \frac{13}{6}c^2 + 8d^2 - cd = 2(c^2 + d^2) + \left(\frac{\sqrt{6}}{6}c - \sqrt{6}d\right)^2 + cd > 2(c^2 + d^2)$, which is in contradiction with $(1)$. Thus, $p = 3$. If $q, r \ge 11$, then note that $\frac{13}{2}\varphi^2(c) \ge \frac{13}{2}\left(\frac{2}{3} \cdot \frac{10}{11} \cdot \frac{18}{19}\right)^2 c^2= \frac{93600}{43681}c^2 > 2c^2$ and $11\varphi^2(d) = 11\left(\frac{2}{3}\right)^2 d^2 = \frac{44}{9}d^2 > 4d^2$. Also note that $\varphi(c)d = \left(\frac{p-1}{p} \cdot \frac{q-1}{q} \cdot \frac{r-1}{r}\right)cd = \varphi(cd)$. So $7\varphi^2(c) - \varphi(cd) + 11\varphi^2(d) > \frac{13}{2}\varphi^2(c)+\frac{1}{2}\varphi^2(c) - \varphi(c)d + 11\varphi^2(d) > 2c^2 + 4d^2 + \frac{1}{2}\varphi^2(c) - \varphi(c)d = 2(c^2 + d^2) + \left(\frac{\sqrt{2}}{2}\varphi(c) - \sqrt{2}d\right)^2 + \varphi(c)d > 2(c^2 + d^2)$, which is in contradiction with $(1)$. Thus, one of $q$ or $r$ must be $7$. WLOG let $q = 7$. Then $c = 3 \cdot 7^y \cdot r^z$ for some positive integers $y, z$. If $x \ge 3$, then $27 \mid \varphi(cd), \varphi^2(d), 2d^2$. Taking both sides of $(1)$ modulo $27$, we have $7\varphi^2(c) \equiv 2c^2 \pmod{27} \iff 144 \cdot 7^{2y-1} \cdot (r-1)^2 \cdot r^{2z-2} \equiv 18 \cdot 7^{2y} \cdot r^{2z} \pmod{27} \iff 72(r-1)^2 \equiv 63r^2 \pmod{27} \iff 8(r-1)^2 \equiv 7r^2 \pmod{3} \iff r^2 - 16r + 8 \equiv 0 \pmod{3} \iff (r - 8)^2 \equiv 56 \equiv 2 \pmod{3}$, which is impossible. Hence, $x = 2$, and $d = 9$. Plugging $c = 3 \cdot 7^y \cdot r^z$ and $d = 9$ into $(1)$, we get $$144 \cdot 7^{2y-1} \cdot (r-1)^2 \cdot r^{2z-2} - 108 \cdot 7^{y-1} \cdot (r-1) \cdot r^{z-1} + 396 = 18 \cdot 7^{2y} \cdot r^{2z} + 162$$$$8 \cdot 7^{2y-1} \cdot (r-1)^2 \cdot r^{2z-2} - 6 \cdot 7^{y-1} \cdot (r-1) \cdot r^{z-1} + 13 = 7^{2y} \cdot r^{2z}$$If $y \ge 2$, then $7 \mid 13$, contradiction. Hence, $y = 1$. Plugging this in, 
$$56(r-1)^2 r^{2z-2} - 6(r-1)r^{z-1} + 13 = 49 \cdot r^{2z}$$If $z \ge 2$, then $r \mid 13 \implies r = 13$. However, $r \equiv 3 \pmod{4}$, contradiction. Hence, $z = 1$. Plugging this in,
$$56(r-1)^2 - 6(r-1) + 13 = 49r^2$$$$7r^2 - 118r + 75 = 0$$Note that the discriminant $118^2 - 4 \cdot 7 \cdot 75 = 11824$, and $108^2 < 11824 < 109^2$, so the discriminant is not a perfect square. Hence, the quadratic equation has no integer solutions. Therefore, there are no solutions where $d = p^x$ where $x \ge 2$. Thus, there are no solutions to this subsubcase. Thus, the only solution to this subcase is $(c, d) = (15, 3)$. Thus, the only solution to this case is $(a, b) = (2c, 2d) = (2 \cdot 15, 2 \cdot 3) = (30, 6)$.