Repeated Substitution

The equation given to us is $2x^{2}-3y^{2}=2014.$ Now, $2x^{2}$ =even and $2014=$ even.Thus, $3y^{2}=$ even $\Longrightarrow y^{2}=$ even $\Longrightarrow y=$ even.Let us say that $x=(2b+1),y=2a.$ Plugging that into the equation given,we get $:2(2b+1)^{2}-3(4a^{2})=2014.$ Simplifying gives $8b^{2}+2+8b-12a^{2}=2014.$ This gives, $8b(b-1)-12a^{2}=2012\Longrightarrow 2b(b-1)-3a^{2}=503$ (dividing by 4).This gives that $3a^{2}=$ odd. $\Longrightarrow a=$ odd.Taking $a=2c+1$ gives $2b^{2}+2b-12c^{2}-3-12c=503.$ $\Longrightarrow 2(b^{2}+b-6c^{2}-6c=506.$ This simplifies to $b(b+1)-6c(c+1)=253.$ But this can't happen as $b(b+1)=$ even.Because the product of two consecutive numbers is always even.Obviously, $6c(c+1)=$ even.But even-even=even.Thus,the answer is $0.$

$2{ x }^{ 2 }-3{ y }^{ 2 }=2014\\ \frac { -3 }{ 2 } { y }^{ 2 }=1007-{ x }^{ 2 }\\ { y }^{ 2 }=(1007-x^{ 2 })(\frac { -2 }{ 3 } )\\ \therefore \frac { (x^{ 2 }-1007)(2) }{ 3 } ={ y }^{ 2 }\\ 2\quad \equiv \quad 2\quad (mod\quad 3)\\ \therefore \quad x^{ 2 }\quad \equiv \quad 2\quad (mod\quad 3)\\ But\quad this\quad is\quad impossible.\quad x^{ 2 }\quad can\quad only\quad be\quad identical\quad to\quad 0\quad or\quad 1.\\ Therefore,\quad there\quad are\quad no\quad such\quad values.$

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