Saturday, 21 July 2018

elementary number theory - Difficult diophantine equation



Solve for integers:



4n4+7n2+3n+6=m3.




Hi this is a problem from an Bulgarian olympiad for which I have no idea how to solve.



I figured out using wolfram alpha that 16m347 must be a square number.



I would appreciate any solutions. Thank you in advance!


Answer



The idea is going modulo 9.



Indeed, we will prove that 4n4+7n2+3n+6 leaves only remainders 2,5,6 modulo 9. None of these are cubes modulo 9(only 0,1,8 are), completing the proof that no such integers n,m exist.




For this, we note that if n0(mod3) then 4n4+7n2+3n+66(mod9).



If n1(mod3) then 4n4+7n2+3n+62(mod9).



Finally, if n1(mod3) then 4n4+7n2+3n+65(mod9).


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