Try this beautiful problem from the Pre-RMO, 2017, Question 23, based on Solving Equation.
Suppose an integer r, a natural number n and a prime number p satisfy the equation \(7x^{2}-44x+12=p^{n}\). Find the largest value of p.
Equation
Algebra
Integers
Answer: is 47.
PRMO, 2017, Question 23
Higher Algebra by Hall and Knight
\(7x^{2}-44x+12=p^{n}\)
or, \(7x^{2}-42x-2x+12=p^{n}\)
or, \((7x-2)(x-6)=p^{n}\)
or, \(7x-2=p^{\alpha}\), \(x-6=p^{\beta}\)
or, \((7x-2)-7(x-6)\)=\(p^{\alpha}-7p^{\beta}\)
\(40=p^{\alpha}-7p^{\beta}\)
If \({\alpha}, {\beta}\) are natural numbers, p is a divisor of 40
or, p=2 or 5
If p=2, 40=\(2^{\alpha}-7(2^{\beta})\) or, \((2^{3})(5)\)=\(2^{\alpha}-7(2^{\beta})\)
or, \({\beta}\)=3 and \(2^{\alpha}\)=40+56
or, \({\alpha}\) not an integer hence not possible
If p=5 then 40=\(5^{\alpha}-7(5^{\beta})\)
or, \((2)^{3}(5)=5^{\alpha}-7(5^{\beta})\)
or, \({\beta}=1\) and \(5^{\alpha}\)=40+35
or, \({\alpha}\) not an integer hence not possible
so \({\beta}\)=0 or, \(p^{\alpha}\)=47
or, p=47 and \({\alpha}\)=1.
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