Try this beautiful problem from the PRMO, 2017 based on Real Numbers and Integers.
PRMO, 2017, Question 2
Suppose a, b are positive real numbers such that \(a(a)^\frac{1}{2}+b(b)^\frac{1}{2}=183\), \(a(b)^\frac{1}{2}+b(a)^\frac{1}{2}\)=182, find \(\frac{9(a+b)}{5}\).
Real Numbers
Algebra
Integers
Answer: 73
Elementary Algebra by Hall and Knight
here \(a(a)^\frac{1}{2}+b(b)^\frac{1}{2}=183\) and \(a(b)^\frac{1}{2}+b(a)^\frac{1}{2}\)=182
This equation gives integer solutions then a and b must be squares.
Let a=\(P^{2}\)and b=\(Q^{2}\)
\(\Rightarrow a(a)^\frac{1}{2}+b(b)^\frac{1}{2}=P^{3}+Q^{3}=183\) is first equation
and \(a(b)^\frac{1}{2}+b(a)^\frac{1}{2}\)=\(P^{2}Q+Q^{2}P\)
=\(PQ(P+Q)\)=182 is second equation
now first equation +3 second equation gives
\(P^{3}+Q^{3}+3PQ(P+Q)\)=183+\(3 \times 182\)=729
\(\Rightarrow (P+Q)^{3}=9^{3}\)
\(\Rightarrow(P+Q)=9\)
second equation gives PQ(P+Q)=182
\(\Rightarrow PQ=\frac{182}{9}\)
then \(a+b=P^{2}+Q^{2}=(P+Q)^{2}-2PQ\)=\(\frac{365}{9}\)
\(\frac{9(a+b)}{5}\)=\(\frac{9}{5} \times \frac{365}{9}\)=73.
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