Note that when $x$ is odd , $\varphi(4x)=2\varphi(x) , \varphi(2x)=\varphi(x)$ Let $2a=a'$ and $2b=b'$ the equation becomes$$14\varphi^2(a')-\varphi(a'b')+22\varphi^2(b')=a'^2+b'^2$$Which is just the even solutions to this.

@above yes but unfortunately that is only when both a and b is odd.

@above the question stated that $a,b$ are odd integers.

The above problem, i haven't done this problem, but here is the way i think it will give the solution a,b are odd, so a^2+b^2 is divisibled by 4 but not 8. Then LHS also divisibled by 4 but not 8 So we have some idea here: +) phi(a), phi(b) is divisibled by 2 for all a,b>_3. So phi(ab) is divisibled by 8 But this mean phi(a) or phi(b) is divisibled by 4. Assume phi(b) is divisibled by 4 then phi(a) is not divisibled by 2, a contradiction So this mean a or b must <3 and we will solve the small case, which m think is hard cuz it's Chinese's problem

Looks like if we check $deg_2$ then we can prove, that $deg_2(a)=deg_2(b)$ And for such case we will have that $a$ or $b$ has only one prime divisor, not equal $2$

RagvaloD wrote: Looks like if we check $deg_2$ then we can prove, that $deg_2(a)=deg_2(b)$ And for such case we will have that $a$ or $b$ has only one prime divisor, not equal $2$ what is deg(a) ?

laikhanhhoang_3011 wrote: RagvaloD wrote: Looks like if we check $deg_2$ then we can prove, that $deg_2(a)=deg_2(b)$ And for such case we will have that $a$ or $b$ has only one prime divisor, not equal $2$ what is deg(a) ? $deg_p(n)=m$ means that $p^m|n, p^{m+1} \not | n$

RagvaloD wrote: laikhanhhoang_3011 wrote: RagvaloD wrote: Looks like if we check $deg_2$ then we can prove, that $deg_2(a)=deg_2(b)$ And for such case we will have that $a$ or $b$ has only one prime divisor, not equal $2$ what is deg(a) ? $deg_p(n)=m$ means that $p^m|n, p^{m+1} \not | n$ ah ok, i often call that v_2(a) instead deg2(a)

Henry_2001 wrote: Determine all the pairs of positive odd integers $(a,b),$ such that $a,b>1$ and $$7\varphi^2(a)-\varphi(ab)+11\varphi^2(b)=2(a^2+b^2),$$where $\varphi(n)$ is Euler's totient function. redacted due to use of wrong notation(I assumed $\phi^2(a)$ as $\phi(\phi(a))$ instead of $\phi(a)^2$

@above we could only prove that a=p^x or b=p^y, not both yet.

Still no solutions here, so I will show that at least one of numbers $a,b$ is a power of prime. Let us denote the next variables: $$L=7\varphi^2(a)-\varphi(ab)+11\varphi^2(b),$$$$R=2(a^2+b^2).$$As $$7\varphi^2(a)-\varphi(ab)+11\varphi^2(b)\equiv 4 \mod 8,$$we have that $\omega(ab)\geq 2$. Suppose both $a$ and $b$ have at least two prime divisors in their factorisation, so $\varphi^2(a), \varphi^2(b)$ are divisible by 8 and $\varphi(ab)\equiv 4 \mod 8$, therefore $\omega(ab)\leq 2$, so $a=p^{k+1}q^{l+1}, b=p^{m+1}q^{n+1}$ for some prime $p<q$ and nonnegative integers $k,l,m,n$. WLOG, $l\leq n$. If $l<n$ then $v_q(L)=2l$ but $v_q(R)=2l+2$ - contradiction. Hence $l=n$. If $p\not | q-1$ then $k=m$ by similar reasons, but in this case 18 is divisble by $pq$ - contradiction again. Therefore $p|q-1$. Let us notice that $p-1|L$, so $R\equiv 0 \mod p-1$. On the other side $R\equiv 4q^{2l+2} \mod p-1$, so $p-1|4$. If $p=5$ then $\varphi(ab)$ is divisible by 8, so by contradiction $p=3$. As $3|p-1$ we have $q\geq 7$. Further, $$\frac{\varphi(a)}{a},\frac{\varphi(b)}{b}\geq \frac 23 \cdot \frac{q-1}{q} \geq \frac 23 \cdot \frac 67 = \frac 47.$$By AG-GM $$\frac 27 a^2 + \frac{78}{49} b^2 > \frac{a^2}4+b^2 \geq ab ,$$so we have the next chain of inequalities $$ L = 7\varphi^2(a)-\varphi(ab)+11\varphi^2(b) > \frac {16}7 a^2 + \frac {176}{49}b^2 - ab =$$$$ = 2(a^2+b^2) + \left (\frac 27 a^2 + \frac{78}{49} b^2 - ab \right) > 2(a^2+b^2)=R,$$which contradicts given equation.

Using the fact that $\varphi(n) \leq n$ for all positive integers $n$, the equation $(1) \;\; 7\varphi^2(a) - \varphi(ab) + 11\varphi^2(b) = 2(a^2 + b^2)$ give us (since $a,b>1$ yields $\varphi(ab)>1$) $2(a^2 + b^3) < 7a + 11b - 1$, i.e. $(2) \;\; (4a - 7)^2 + (4b - 11)^2 < 13^2$, which implies $a<5$ and $b<6$, Consequently, since $a,b>1$ are both odd integers, the only possible solutions of equation (1) are $(a,b) = (3,3)$ and $(a,b) = (3,5)$. Checking neither of these two pairs are solutions of equation (1) since the both give $LHS<RHS$ in equation (1). Conclusion: Equation (1) has no solution when $a,b>1$ are odd integers.

Solar Plexsus wrote: Using the fact that $\varphi(n) \leq n$ for all positive integers $n$, the equation $(1) \;\; 7\varphi^2(a) - \varphi(ab) + 11\varphi^2(b) = 2(a^2 + b^2)$ give us (since $a,b>1$ yields $\varphi(ab)>1$) $2(a^2 + b^3) < 7a + 11b - 1$, i.e. $(2) \;\; (4a - 7)^2 + (4b - 11)^2 < 13^2$, which implies $a<5$ and $b<6$, Consequently, since $a,b>1$ are both odd integers, the only possible solutions of equation (1) are $(a,b) = (3,3)$ and $(a,b) = (3,5)$. Checking neither of these two pairs are solutions of equation (1) since the both give $LHS<RHS$ in equation (1). Conclusion: Equation (1) has no solution when $a,b>1$ are odd integers. $\varphi^2(a)$ means $\varphi(a)^2$ here: so $(15,3)$ is a solution.

I thought $\varphi^2(n) = \varphi(\varphi(n))$. Thus , since $\varphi^2(n) = [\varphi(n)]^2$, my solution of the equation is obviously incorrect.

Hint: Using the fact that Euler's totient function is bigger than n-n^(1/2), you can eliminate big values of a and b easily.

@above, it phi(n) is smaller than n-n^(1/2), not bigger.

Justanaccount wrote: @above, it phi(n) is smaller than n-n^(1/2), not bigger. Opps ,sorry.

It can easily be shown that if $(a, b)$ is a solution and $p^2$ divides both $a$ and $b$, where $p$ is a prime number, then $\left(\frac{a}{p}, \frac{b}{p}\right)$ is also a solution. Thus, we can assume that $\nu_p (gcd(a, b)) \le 1$ for all primes $p$. Replacing $a$ and $b$ by $c$ and $d$ respectively, this is just Subcase 2.2 to my solution here. Thus, the only solution to this equation such that $\nu_p (gcd(a, b)) \le 1$ for all primes $p$ is $(a, b) = (15, 3)$. Note that 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 prime that divides both $15$ and $3$ is $3$, all solutions to the equation can be expressed in the form $\boxed{(15 \cdot 3^x, 3 \cdot 3^x)}$, where $x$ is a nonnegative integer.