Determine all pairs of positive integers for which is a perfect square.

Indian National Mathematical Olympiad 1992

Number Theory

Easy

An Excursion in Mathematics

Do you really need a hint? Try it first!

First consider . In this case the equation can be rewritten as . Hence, solving this case is equivalent to looking for powers of 2 that differ by 2.

For , consider residues modulo 3. Note that a perfect square cannot have remainder 2 upon division by 3.

Note that . From hint 3, it follows that has to be even. Writing we get .

From hint 3, we may write and for some and . This gives . As the RHS is not divisible by 3, has to be 0. Thus, . If then . Otherwise, the LHS is 0 modulo 4, i.e. . Hence for some . Now we again write . Again, we have two powers of 2 differing by 2. It is easy to see that (in case you have not noticed already) only such powers are 2 and 4. Hence .

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