I cant understand solution of $a^2 + b^2 = 2020$ and $LCM(a,b)=336$

Question

What are the ordered pairs $(a,b) $ in $\mathbb N×\mathbb N$, satisfying the following:

$$a^2+b^2=2020.$$

$$LCM(a,b)=336.$$

THANK YOU.

Answer 1

Since $a,\,b$ must be factors of $336=2^4\times3\times7$ that square to $\le2020<45^2$, they must each be elements of $S:=\{1,\,2,\,3,\,4,\,6,\,7,\,8,\,12,\,14,\,16,\,21,\,24,\,28,\,42\}$. A choice of $a\in S$ determines $b=\sqrt{2020-a^2}$, which either is or isn't an element of $S$. You can easily verify the only solution is $a=16,\,b=42$ or vice versa. We must double-check $16=2^4,\,42=2\times3\times7$ have lcm $2^4\times3\times7$, as indeed they do.

Answer 2

WLOG $(a,b)=d$

and $$\dfrac aA=\dfrac bB=d\implies(A,B)=1$$

$$\dfrac{2020}{336^2}=\dfrac{A^2+B^2}{A^2B^2}$$

$$505A^2B^2=168^2(A^2+B^2)$$

So, $168^2=(505A^2B^2)/(A^2+B^2)$

So, $\dfrac{5\cdot101}{A^2+B^2}$ must be perfect square

which is possible only if $A^2+B^2=(2^2+1^2)(10^2+1^2)$

Use https://proofwiki.org/wiki/Brahmagupta-Fibonacci_Identity