Number Theory

[Naheed]

If 4ab is divisible by (a+b)² then prove that, a=b .

[tanmoy]

See that $4ab=(a+b)^{2}-(a-b)^{2}$.So,$(a+b)^{2}$ divides $(a-b)^{2}$.But $(a-b)^{2}<(a+b)^{2}$ which implies that $(a-b)^{2}=0$.$\therefore$ $a=b$.
[$\text{Please use latex in writing equations}$]

[Naheed]

Thanks a lot! Thought it harder. (a-b)²= 0--that's what I didn't get at first.

[Naheed]

But if b is negative, (a-b)² >(a+b)² .

[nahin munkar]

If b is negative, then your question will not be valid anymore.Because,if b negative, let,b=$(-1)*c$.(Here c is absolute value of b).Replace $(-1)*c$ at the place of b.Then,it will be clear to u .So,a condition should be given that,either (a,b) is positive or (a,b) is negative.not one of (a,b) is positive or negative.

[Tariq Hasan Rizu]

According to AM–GM inequality,
$\frac {a+b}{2}$ $\geq\sqrt{ab}$
$\Rightarrow$ $\frac{(a+b)^{2}}{4}$ $\geq ab$
$\Rightarrow (a+b)^{2}\geq 4ab$
$\because (a+b)^{2}\mid 4ab$
$\Rightarrow (a+b)^{2}\leq 4ab$
$\Rightarrow (a+b)^{2}=4ab$
$\therefore (a-b)^{2}=0$
$\therefore a=b$.