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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)

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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