Solving $a^2+b^2=2020, \text{lcm}(a,b)=336$ for $a$ and $b$

Question

What are the ordered pairs $(a,b)\in \Bbb{N}\times \Bbb{N}$ satisfying the following: $$a^2+b^2=2020.$$ $$\text{lcm}(a,b)=336.$$

THANK YOU.

Answer 1

Hint:

Clearly $a,b$ must be even(why?)

$(a,b)$ must divide $(2020,336)=4$

Check for the positive integer solutions for the two cases

Answer 2

Let $(a,b) = n$, $a = pn$ and $b = qn$. Then $$ a^2 + b^2 = (p^2 + q^2)n^2 = 2020. $$ Thus $n^2$ divides $2020$, but as $2020 = 2^2\cdot5\cdot 101$ contain only one square in its prime decomposition there are only two options: $(a,b) = 1$ and $(a,b) = 2$. Then using the fact that $ab = (a,b)[a,b]$ we may easily check both these two options.

If $(a,b) = 1$ then $ab = 336$ and we get $(a+b)^2 = a^2 + b^2 + 2ab = 2692$ which is not perfect square.

Thus $(a,b) = 2$, then $ab = 2\cdot 336 = 672$, and we get $$ (a+b)^2 = 2020 + 2\cdot672 = 58^2 \\ (a-b)^2 = 2020 - 2\cdot672 = 26^2 $$ which gives two solutions $a = 42$, $b=16$ or vice cersa.

Answer 3

Note that $2020=4\cdot505$ and $336=16\cdot21$. Since both $2020=a^2+b^2$ and $336=\rm{lcm}(a,b)$ are even, $a$ and $b$ must both be even. Writing $a=2\alpha$ and $b=2\beta$, so that $\alpha^2+\beta^2=505$, we see that one of $\alpha,\beta$ is even and the other is odd. Let's assume $\alpha$ is even and $\beta$ is odd. Then we must have $a=2\alpha=16\alpha'$ with $\alpha'$ odd, in order for $16$ be the largest power of $2$ to divide $\rm{lcm}(a,b)=\rm{lcm}(2\alpha,2\beta)$ with odd $\beta$. But since $48^2=2304\gt2020$, we can only have $\alpha'=1$. Hence $a=16$, which implies $b=\sqrt{2020-256}=42=2\cdot21$. Thus the only solutions in positive integers are $(a,b)=(16,42)$ and $(a,b)=(42,16)$.

Answer 4

LCM is $336 = 2^4*3*7$. So one is divisible by 3 and the other may or may not be. One is divisible by 7 and the other may or may not be. One is divisible by 16 and the other, if divisible by 2, is divisible up to but not more than $2^4$

If $d = \gcd(a,b)$ then $d^2|a^2 + b^2 = 4^2*5*101$-- which is divisible by $2^2$ but no higher power, and not divisible by $3$ or $7$. So that means $7$ divides one but not the other, $3$ divides one and not the other, and although $16$ divides one the greatest power of $2$ that divides the other is $2$. Furthermore, we know $2$ divides the other. (If $p|a$ and $p|a^2 + b^2$ then $p|b^2 = (a^2 +b^2) - a^2$)

So we know (wolog) $a = 2a'$ and $b = 16b'$ (or vice versa), $a',b'$ odd. and $a'^2 + 64b'^2 = 505$.

If $7|b'$ we have $64b'^2 \ge 64*49 > 505$ so $7|a'$. Also if $3|b'$ we have $64b'^2 \ge 64*9 > 505$. So $3,7|a'$ and we have:

$a'=21a"$ and $441a"^2 + 64b'^2 =505$.

Okay, doesn't take much to figure out that $a" =1; b'=1$ is the only solution. So $a = 42$ and $b=16$ (or vice versa) and $a^2 + b^2 = 1764 + 256 = 2020$.

So the ordered pairs are $(42,16)$ and $(16,42)$.