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Albert1
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$x,y \in N$
Prove :$x^2+y^2=1992$ has no solution
Prove :$x^2+y^2=1992$ has no solution
Albert said:$x,y \in N$
Prove :$x^2+y^2=1992$ has no solution
The equation x²+y²=1992 represents a circle with its center at the origin and a radius of √1992.
This equation has no solution because the sum of two squares cannot equal a non-square number.
We can prove that there is no solution by using the Pythagorean theorem, which states that in a right triangle, the sum of the squares of the two shorter sides is equal to the square of the longest side.
In this case, we can see that 1992 is not a perfect square, so there is no way for x²+y² to equal 1992.
No, we cannot find a solution to this equation using complex numbers because complex numbers include a real and an imaginary part, neither of which can be squared to equal 1992.
Yes, we can also prove that there is no solution by graphing the equation and seeing that it does not intersect with the x or y axis. This means that there are no real number solutions for x and y that will make the equation true.