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 16⋅m3−47 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 n≡0(mod3) then 4n4+7n2+3n+6≡6(mod9).
If n≡1(mod3) then 4n4+7n2+3n+6≡2(mod9).
Finally, if n≡−1(mod3) then 4n4+7n2+3n+6≡5(mod9).
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