Prove that there are no solutions in integers x and y to the equation

2x^2 + 5y^2 = 14

Thanks in advance

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- Apr 10th 2006, 08:52 PMjzonHow does this go?
Prove that there are no solutions in integers x and y to the equation

2x^2 + 5y^2 = 14

Thanks in advance - Apr 11th 2006, 12:05 AMticbolQuote:

Originally Posted by**jzon**

y can only be 0 or +,-1, because if y were |2| or greater, then 5y^2 is already 20 or more---which is greater than 14.

If y=0, then x^2 = 7, and x is not an integer.

If y= |1|, then x^2 = 9/2, and x is not an integer.

x can only be 0, |1|, or |2|. If x were |3| or more, then 2x^2 is already 18 or more.

If x=0, y^2 = 14/5 and y is not an integer.

If x=|1|, y^2 = 12/5 and y is not an integer.

If x=|2|, y^2 = 16/5 and y is not an integer.

Therefore, there is no solution in x and y as integers.