Answer
In order to prove that there exist no solutions in the integers x and y to the equation.
2x2+5y2=14
a step by step solution is given below..
Work Step by Step
Since -a^{2}=a^{2} for all integers and since the equation contains squares of the unknown integers, it is safe to assume that the integers x and y are nonnegative.
Since the squares are never negative and since the equation contains no differences (or negative sign), we only need to check all integers with squares smaller than 14. The only integers with squares smaller than 14 are 1, 2 and 3.
1^{2}=1
2^{2}=4
3^{2}=9
Thus x2 and y2 can take on the values of 1, 4 and 9. let us check if there is a possible combination make the equation true.
2(1)+5(1)=7 \neq 14
2(1)+5(4)=22 \neq 14
2(1)+5(9)=47 \neq 14
2(4)+5(1)=13 \neq 14
2(4)+5(4)=28 \neq 14
2(4)+5(9)=53 \neq 14
2(9)+5(1)=23 \neq 14
2(9)+5(4)=38 \neq 14
2(9)+5(9)=63 \neq 14
None of the possible combinations of x^{2} square and y^{2} square made the equation true and hence there are no solutions in integers x and y
