These Are Not The Solutions You Are Looking For

$x^{2}=(x-y)(3x+3y+1)$ and $y^{2}=(x-y)(2x+2y+1)$ .

Now, $x-y=gcd(x^{2},y^{2}=gcd(x,y)^{2}$ .Let $gcd(x,y)=d$ .So $x=db$ and $y=da$ where $gcd(a,b)=1$

Also since $3(x+y)+1)$ and $2(x+y)+1$ are relatively prime, they must be perfect squares.

Again, $d^{2}b^{2}=x^{2}=(x-y)(3x+3y+1)=d^{2}(3x+3y+1)$ .So $3x+3y=1=b^{2}$

Similarly, $2x+2y+1=a^{2}$ .

Now, $d^{2}=x-y=db-da$ and hence d=b-a

$3a^{2}-2b^{2}=1$ and its solutions can be obtained from $(\sqrt{3}+\sqrt{2})^{2n+1}=a_{n}\sqrt{3} +b_{n}\sqrt{2}$ Now, $a_{n+1}\sqrt{3}+b_{n+1}\sqrt{2}=(a_{n}\sqrt{3}+b_{n}\sqrt{2})(5+2\sqrt{6})$

We get $a_{n+1}=5a_{n}+4b_{n},b_{n+1}=6a_{n}+5b_{n},a_{1}=1,b_{1}=1$

From this recurrence obtain the first 4 solutions and compute the sum.