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# Understand the problem

Positive real numbers $a$ and $b$ have the property that$$\sqrt{\log{a}} + \sqrt{\log{b}} + \log \sqrt{a} + \log \sqrt{b} = 100$$and all four terms on the left are positive integers, where log denotes the base 10 logarithm. What is $ab$? $\textbf{(A) } 10^{52} \qquad \textbf{(B) } 10^{100} \qquad \textbf{(C) } 10^{144} \qquad \textbf{(D) } 10^{164} \qquad \textbf{(E) } 10^{200}$

# 2019 AMC 12A Problems/Problem 15

##### Topic
logarithm, diophantine equation
Medium
##### Suggested Book
Mathematical Circles (Russian Experience)

# Start with hints

Do you really need a hint? Try it first!

Given both $\sqrt {\log a} , \sqrt {\log b}$ are positive integers .  $\Rightarrow$ both $\log a ,\log b$ are perfect squares . similarly , both $\log {\sqrt a} , \log {\sqrt b}$ are positive integers. $\Rightarrow$ both $a ,b$ are perfect squares .
so $$\log a = m^2 , \log b = n^2$$ where $m,n \in {Z^+}$ $\Rightarrow a= 10^{m^2} , b= 10^{n^2}$  and as both $a, b$ are perfect squares  $\\ \Rightarrow 10^{m^2} ,10^{n^2}$ are both perfect squares i.e $10^{m^2} = p^2,10^{n^2} =q^2$ , where $p,q \in {Z^+}$ . $\\ \Rightarrow \frac {m^2}{2} , \frac {n^2}{2}$ are integers $\\ \Rightarrow 2|m^2 , 2|n^2$ $\\ \Rightarrow 2|m, 2|n$ (as $2$ is a prime number ) $\\$ so now $a= 10^{4x^2} , b= 10^{4y^2}$ can be put in the original equation , where  $x,y \in {Z^+}$ .
Now to get the solution from the derived equation i.e. $2x + 2y + 2x^2 + 2y^2 =100$  multiply both the sides by $2$ and then add $2$ in both sides to arrive at $(2x+1)^2 + (2y+1)^2 = 202$ .
Now use trial and error method to express $202$ as a sum of two odd perfect squares . Finally the only way i.e. $9^2 + 11^2 = 202$ .  So without loss of generality it can be written that $(2x+1) = 9 , (2y+1)= 11$ So $a= 10^{64} , b= 10^{100}$  and $ab = 10^{164}$

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