Determine all the pairs of positive integers $(a,b),$ such that $$14\varphi^2(a)-\varphi(ab)+22\varphi^2(b)=a^2+b^2,$$ where $\varphi(n)$ is Euler's totient function.
1. **Initial Assumptions and Simplifications:**
   - Given the equation:
     \[
     14\varphi^2(a) - \varphi(ab) + 22\varphi^2(b) = a^2 + b^2,
     \]
     where \(\varphi(n)\) is Euler's totient function.
   - If \((a, b)\) is a solution and there exists a prime \(p\) such that \(p^2 \mid a, b\), then:
     \[
     14\varphi^2\left(\frac{a}{p}\right) - \varphi\left(\frac{a}{p} \cdot \frac{b}{p}\right) + 22\varphi^2\left(\frac{b}{p}\right) = \left(\frac{a}{p}\right)^2 + \left(\frac{b}{p}\right)^2.
     \]
     This implies \(\left(\frac{a}{p}, \frac{b}{p}\right)\) is also a solution.

2. **Parity Considerations:**
   - Since \(\varphi(n) < n\) for all \(n \geq 2\) and \(\varphi(n)\) is even for all \(n \geq 3\), we can assume \(\nu_p(\gcd(a, b)) \leq 1\) for all primes \(p\).
   - If \(ab \leq 2\), then \(a, b \leq 2\), leading to a contradiction since \(14\varphi^2(a) + 22\varphi^2(b) \geq 36 > 2^2 + 2^2 > a^2 + b^2 - \varphi(ab)\).
   - Therefore, \(ab \geq 3\) and \(\varphi(ab)\) is even. Since \(14\varphi^2(a) - \varphi(ab) + 22\varphi^2(b)\) is even, \(a^2 + b^2\) must be even, implying \(a\) and \(b\) have the same parity.

3. **Case 1: Both \(a\) and \(b\) are odd:**
   - \(a^2 + b^2 \equiv 2 \pmod{4}\), 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 \(a, b \geq 3\), then \(\varphi(ab) \equiv 0 \pmod{4}\), leading to a contradiction.

4. **Subcase 1.1: Both \(a\) and \(b\) are equal to 1:**
   - This leads to \(35 = 2\), which is a contradiction.

5. **Subcase 1.2: Exactly one of \(a\) or \(b\) is equal to 1:**
   - WLOG, let \(b = 1\). Then:
     \[
     14\varphi^2(a) - \varphi(a) + 22 = a^2 + 1.
     \]
   - If \(8 \mid \varphi(a)\), then \(22 \equiv a^2 + 1 \pmod{8} \implies a^2 \equiv 5 \pmod{8}\), which is impossible.
   - Therefore, \(\nu_2(\varphi(a)) \leq 2\), and \(a\) must have at most 2 odd prime factors.
   - If \(a\) has 2 odd prime factors \(p\) and \(q\), then:
     \[
     14\varphi^2(a) = 14\left(\frac{p-1}{p} \cdot \frac{q-1}{q}\right)^2 a^2 \geq 14\left(\frac{2}{3} \cdot \frac{4}{5}\right)^2 a^2 = \frac{896}{225}a^2 > 3a^2 > a^2 + a,
     \]
     leading to a contradiction.
   - If \(a\) has 1 odd prime factor \(p\), then:
     \[
     14\varphi^2(a) = 14\left(\frac{p-1}{p}\right)^2 a^2 \geq 14\left(\frac{2}{3}\right)^2 a^2 = \frac{56}{9}a^2 > 3a^2 > a^2 + a,
     \]
     leading to a contradiction.

6. **Subcase 1.3: \(a\) and \(b\) are both not equal to 1:**
   - \(a, b \geq 3\), hence \(ab\) must be the power of an odd prime.
   - Since \(\nu_p(\gcd(a, b)) \leq 1\) for all primes \(p\), either \((a, b) = (p^k, p)\) or \((p, p^k)\) for some positive integer \(k\) and odd prime \(p\).
   - In both cases, \(p-1 \mid \varphi(a), \varphi(b), \varphi(ab)\), so \(p-1 \mid a^2 + b^2 = p^{2k} + p^2 = p^2(p^{2k-2} + 1) \implies p-1 \mid p^{2k-2} + 1 \implies p-1 \mid 2 \implies p = 3\).
   - 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 when both \(a\) and \(b\) are odd.

7. **Case 2: Both \(a\) and \(b\) are even:**
   - 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 \(14 - 2 + 22 = 8 \implies 34 = 8\), contradiction.
   - If \(c \geq 2\), 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 \(a, b > 2\).

8. **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 \(\nu_2(14\varphi^2(a) + 22\varphi^2(b)) \geq 3\). 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 \(l \geq 3\) 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 \(\implies m = 1\). So \(ab = 8\).
   - However, since \(a, b > 2\), \(ab \geq 9\), contradiction.
   - Thus, there are no solutions for this subcase.

9. **Subcase 2.2: \(c\) and \(d\) have the same parity:**
   - Then both \(c\) and \(d\) must be odd.
   - The equation becomes:
     \[
     14\varphi^2(2c) - \varphi(4cd) + 22\varphi^2(2d) = 4(c^2 + d^2).
     \]
   - Simplifying, we get:
     \[
     7\varphi^2(c) - \varphi(cd) + 11\varphi^2(d) = 2(c^2 + d^2).
     \]
   - 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:
     \[
     7\varphi^2(c) \geq 7\left(\frac{2}{3}\right)^2c^2 = \frac{28}{9}c^2 > \frac{5}{2}c^2,
     \]
     and:
     \[
     11\varphi^2(d) \geq 11\left(\frac{2}{3}\right)^2d^2 = \frac{44}{9}d^2 > \frac{5}{2}d^2.
     \]
   - Then:
     \[
     7\varphi^2(c) - \varphi(cd) + 11\varphi^2(d) > \frac{5}{2}(c^2 + d^2) - cd = 2(c^2 + d^2) + \frac{1}{2}(c-d)^2 \geq 2(c^2 + d^2),
     \]
     which is in contradiction with the equation. 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 \(\nu_2(\varphi(cd)) \geq 2\).

10. **Subsubcase 2.2.1: \(cd\) has exactly 2 distinct prime factors:**
    - Let those prime factors be \(p\) and \(q\) such that \(p < q\).
    - If \(p \geq 5\), then:
      \[
      7\varphi^2(c) \geq 7\left(\frac{4}{5} \cdot \frac{6}{7}\right)^2 c^2 = \frac{576}{125}c^2 > \frac{5}{2}c^2,
      \]
      and:
      \[
      11\varphi^2(d) \geq 11\left(\frac{4}{5} \cdot \frac{6}{7}\right)^2 d^2 = \frac{6336}{1225}d^2 > \frac{5}{2}d^2.
      \]
    - Then:
      \[
      7\varphi^2(c) - \varphi(cd) + 11\varphi^2(d) > \frac{5}{2}(c^2 + d^2) - cd = 2(c^2 + d^2) + \frac{1}{2}(c-d)^2 \geq 2(c^2 + d^2),
      \]
      which is in contradiction with the equation.

11. **Subsubcase 2.2.2: \(cd\) has exactly 3 distinct prime factors:**
    - Since \(\nu_2(\varphi(cd)) \leq 3\), 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:
      \[
      7\varphi^2(c) - \varphi(cd) + 11\varphi^2(d) \equiv 4 \pmod{8} \implies 7\varphi^2(c) + 11\varphi^2(d) \equiv 4 \pmod{8}.
      \]
    - 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.

12. **Conclusion:**
    - The only solution such that \(\nu_p(\gcd(a, b)) \leq 1\) for all primes \(p\) is \((a, b) = (30, 6)\).
    - If \((a, b)\) is a solution and if a prime \(p\) divides both \(a\) and \(b\), then \((ap, bp)\) is also a solution.
    - Since the only primes that divide both \(30\) and \(6\) are \(2\) and \(3\), all solutions to the equation can be expressed in the form:
      \[
      \boxed{(30 \cdot 2^x \cdot 3^y, 6 \cdot 2^x \cdot 3^y)},
      \]
      where \(x\) and \(y\) are nonnegative integers.