2021/114 solve in integers $3x^2 + 5y^2 = 345$

working in mod 3 we have $5y^2 = 0 \pmod 3$ or $y =0 \pmod 3$

so y = 3a for some a

similarly x = 5 b for som b

so ge get $75 b^2 + 45 y^2 = 345$ 

deviding by 15 we get $5b^2 + 3a^2 = 23$

we need to check for $5b^2 < 23$ or $b <=2$

putting b = 1 we get  $3a^2 = 18$ or $a^2 = 6$ not an integer

b = 2 gives $3a^3 =3$ or a = 1

so we have a= 1 , b= 2 giving x = 10 and y = 3