Number Theory Problem
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Prove that if $(a+b)^2$ divides $4ab$ then $a=b$
Hmm..Hammer...Treat everything as nail
- Anindya Biswas
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Re: Number Theory Problem
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 $(a-b)^2=0\Rightarrow a=b$
$4ab=(a+b)^2-(a-b)^2$
$\Rightarrow (a+b)^2|(a-b)^2$
Since $0\leq (a-b)^2<(a+b)^2$,
The only possible way is $(a-b)^2=0\Rightarrow a=b$
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
— John von Neumann
— John von Neumann
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Re: Number Theory Problem
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 $a=b$
$(a+b)^2|4ab$ 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 $a=b$
Hmm..Hammer...Treat everything as nail
- Mehrab4226
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Re: Number Theory Problem
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,
$(a+b)^2=4ab$
$\therefore a=b$
$(a+b)^2|4ab$ 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,
$(a+b)^2=4ab$
$\therefore a=b$
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré
-Henri Poincaré
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Re: Number Theory Problem
This was overkill for a juniorMehrab4226 wrote: ↑Fri Feb 26, 2021 10:39 pmMy 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,
$(a+b)^2=4ab$
$\therefore a=b$
Hmm..Hammer...Treat everything as nail