Number Theory Problem

For students of class 6-8 (age 12 to 14)
Asif Hossain
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Number Theory Problem

Unread post by Asif Hossain » Fri Feb 26, 2021 7:40 am

Prove that if $(a+b)^2$ divides $4ab$ then $a=b$
Hmm..Hammer...Treat everything as nail

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Anindya Biswas
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Re: Number Theory Problem

Unread post by Anindya Biswas » Fri Feb 26, 2021 10:15 am

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 $(a-b)^2=0\Rightarrow a=b$
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

Asif Hossain
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Re: Number Theory Problem

Unread post by Asif Hossain » Fri Feb 26, 2021 4:06 pm

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 $a=b$
Hmm..Hammer...Treat everything as nail

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Mehrab4226
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Re: Number Theory Problem

Unread post by Mehrab4226 » 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,
$(a+b)^2=4ab$
$\therefore a=b$
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

Asif Hossain
Posts:194
Joined:Sat Jan 02, 2021 9:28 pm

Re: Number Theory Problem

Unread post by Asif Hossain » Sat Feb 27, 2021 11:46 am

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 :lol: :lol: for a junior
Hmm..Hammer...Treat everything as nail

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