For $7(a+b)^2 = 320m$. Find The Value Of $a$ And $b$.

Question

suppose that $a$ and $b$ are natural numbers, $a ≠ 0$ and $b ≠ 0$,

and $lcm(a,b) = m$.

Given that $$7(a+b)^2 = 320m$$

Find the possible values for both $a$ and $b$.

Answer 1

Hint

Define $\gcd(a,b)=d$ and $u={a\over b}$. Since $$dm=ab$$ we obtain $$7(a+b)^2=320{ab\over d}\iff u^2+2u+1={320\over 7d}{u}$$which leads to $$u={160\over 7d}-1\pm\sqrt{\left({160\over 7d}-1\right)^2-1}={160\over 7d}-1\pm 8{\sqrt{400-35d}\over 7d}$$since $u\in\Bbb Q$ and $d\in\Bbb N$, then $400-35d$ must be a perfect square which leaves us with a few options for $d$. The rest is easy.

Answer 2

Let $\gcd(a,b)=d$ let $a=dc$ and $b=de$ thus $$\gcd(c+e,e)=\gcd(c,c+e)=\gcd(c,e)=1$$ and $$7d^{3}(c+e)^{2}=7(a+b)^{2}d=320md=320ab=320d^{2}ce$$. Thus $$7d(c+e)^{2}=320ce$$from here, note that, $$(c+e)^{2}|320=2^{6}*5$$ ,thus only possible choices for $c+e$ are $c+e=2,4,8$ but $$7|ce$$ thus either $c$ or $e$ must be the multiple of 7. By symmetry let ,$c=7k$ thus $$c+e\geq 8$$ thus $$c+e=8$$ then $c=7$ and $e=1$ is forced!!! then $d=5$.Thus only two pairs that satisfy $(a,b)=(35,5),(5,35)$ and we are done.

We don't need the bound for $d$ to do the real solution in much easier way, however bounding $d$ is not a bad idea either. Cheers!! Initially I was lazy to think so just typed it as a hint but here is the actual solution. Remark: Please try it by bounding $d$ and case analysis , it will be more fun!! Note $1\leq d\leq11$

Answer 3

A few properties you could exploit:

• lcm($a$,$b$) divides $ab$
• 7 being prime means either 320 divides by it ($320=2^6\cdot 5$), or $m$ does.
• let $m=7k$ then $a^2+2ab+b^2$ is divisible by $k$ which means it divides $a^2+b^2$
• parity ( mod 2) shows $a\equiv b\bmod 2$
• divisibility by 8 shows $a$ and $b$ must both be odd. ( in fact one must be 3 mod 4)
• etc.