Find $a,b$ such that $a^2+b^2=85113$ and that $\text{lcm}\{a,b\}=1764$ 
Find two positive integers $a,b$ such that $a^2+b^2=85113$ and that $\text{lcm}\{a,b\}=1764$

I have completely no idea where to enter this question.
 A: Note that $\text{lcm}(a,b)=1764=(2\cdot 3\cdot 7)^2$ and $85113=3^2\cdot 7^2\cdot 193$. We may assume that $7^2$ divides $b$ then $a^2=85113-b^2$ is divisible by $7^2$ which means that $7$ divides $a$. Let $A:=a/7$, $B:=b/7$ and solve the reduced diophantine equation
$$A^2+B^2=85113/7^2=1737$$
with $\text{lcm}(A,B)=2^2\cdot 3^2\cdot 7$.
Similarly, we can operate with respect to the factor $3$ and solve the reduced diophantine equation
$$X^2+Y^2=1737/3^2=193$$
where $X:=a/21$ and $Y:=b/21$ with $\text{lcm}(X,Y)=2^2\cdot 3\cdot 7$.
Since $13^2<193<14^2$, now it is easy to find the solutions by hand. Can you take it from here?
A: There is no rigorous way to approach the solution, but these are some tips that might help. 
Let's suppose $a=b$. Then $a = b = \sqrt{85113/2} \approx 206$. Thus the geometric mean of the numbers is around $206$.
Also consider that $85113 \equiv 1 \pmod 4$. This means there is one odd number and one even number.
Finally consider the factorisation of $1764$ as $2^2 \cdot 3^3 \cdot 7^2$.  Which two factors of $1764$ satisfy the two properties above? There are only a few odd factors, which leaves a few cases to check by hand.
