Let p

1

p

2

be odd primes satisfy p

1

≡p

2

≡1(mod 6) and (p

1/

p

2

)=－1

where (p

1/

p

2

) is the Legendre symbol. Let Q be a positive integer such that Q is square free

and it has at least two distinct prime divisors and every prime divisor q of Q satisfies q≡5(mod 6). Using some elementary number theory methods

it is proven that if p

1

≡1(mod 8)

p

2

≡5(mod 8) and Q≡1(mod 4)

 then the equation x

3

+1=2p

1

p

2

Qy

2

has no positive integer solutions (x

y).