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 $ab$? $\textbf{(A) }   10^{52}   \qquad        \textbf{(B) }   10^{100}   \qquad    \textbf{(C) }   10^{144}   \qquad   \textbf{(D) }  10^{164} \qquad  \textbf{(E) }   10^{200}$

Source of the problem

2019 AMC 12A Problems/Problem 15

Topic
logarithm, diophantine equation
Difficulty Level
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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