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Number Theory Problem
Posted: Fri Feb 26, 2021 7:40 am
by Asif Hossain
Prove that if $(a+b)^2$ divides $4ab$ then $a=b$
Re: Number Theory Problem
Posted: Fri Feb 26, 2021 10:15 am
by Anindya Biswas
I am assuming $a,b$ positive integer. Otherwise this claim doesn't work.
$4ab=(a+b)^2-(a-b)^2$
$\Rightarrow (a+b)^2|(a-b)^2$
Since $0\leq (a-b)^2<(a+b)^2$,
The only possible way is $(a-b)^2=0\Rightarrow a=b$
Re: Number Theory Problem
Posted: Fri Feb 26, 2021 4:06 pm
by Asif Hossain
Well i did it in this way
$(a+b)^2|4ab$ implies $(a+b)^2\leq 4ab$ which in turns implies $(a-b)^2 \leq 0$ but $0 \leq (a-b)^2$ so the only possiblity is $(a-b)^2=0$ implying $a=b$
Re: Number Theory Problem
Posted: Fri Feb 26, 2021 10:39 pm
by Mehrab4226
My one is like this:
$(a+b)^2|4ab$ implies,
$(a+b)^2\leq4ab$ $\cdots (1)$
Again by AM-GM we get,
$\frac{a+b}{2}\geq \sqrt{ab}$
Or,$\frac{(a+b)^2}{4}\geq ab$
Or, $(a+b)^2 \geq 4ab$ $\cdots (2)$
(1) and (2) implies,
$(a+b)^2=4ab$
$\therefore a=b$
Re: Number Theory Problem
Posted: Sat Feb 27, 2021 11:46 am
by Asif Hossain
Mehrab4226 wrote: ↑Fri Feb 26, 2021 10:39 pm
My one is like this:
$(a+b)^2|4ab$ implies,
$(a+b)^2\leq4ab$ $\cdots (1)$
Again by AM-GM we get,
$\frac{a+b}{2}\geq \sqrt{ab}$
Or,$\frac{(a+b)^2}{4}\geq ab$
Or, $(a+b)^2 \geq 4ab$ $\cdots (2)$
(1) and (2) implies,
$(a+b)^2=4ab$
$\therefore a=b$
This was overkill
for a junior