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solve in positive integers: 3 \cdot 2^x +4 =n^2

Source: Greece JBMO TST 2019 p2

April 29, 2019
number theoryDiophantine equationpositive integers

Problem Statement

Find all pairs of positive integers (x,n)(x,n) that are solutions of the equation 3ā‹…2x+4=n23 \cdot 2^x +4 =n^2.
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