Prove :x²+y²=1992 ,no solution

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In summary, 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 this by using the Pythagorean theorem and observing that 1992 is not a perfect square. Additionally, we cannot find a solution using complex numbers because neither the real nor imaginary parts can be squared to equal 1992. Another way to prove that there is no solution is by graphing the equation and observing that it does not intersect with the x or y axis, meaning there are no real number solutions for x and y.
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Albert1
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$x,y \in N$

Prove :$x^2+y^2=1992$ has no solution
 
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Albert said:
$x,y \in N$

Prove :$x^2+y^2=1992$ has no solution

if x is odd say 2n + 1 then
$x^2= 4n^2 + 4n + 1 = 4n (n+1) + 1 = 1$ mod 8

and $x^2= 0/4 $ mod 8 if x is even as $x^2+y^2 = 1992$ mod 8 so both x and y are even because if one is odd then it is odd mod 8 and if both are odd it is 2 mod 8

let x = 2a and y = 2b

$x^2+y^2 = 1992$
or $a^2 + b^2 = 498$

so $a^2+b^2= 6 $ mod 8
but from above $a^2+b^2$ mod 8 can be 0 or 1 or 2 or 5

so no solution
 
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FAQ: Prove :x²+y²=1992 ,no solution

What does the equation x²+y²=1992 represent?

The equation x²+y²=1992 represents a circle with its center at the origin and a radius of √1992.

Why is there no solution to the equation x²+y²=1992?

This equation has no solution because the sum of two squares cannot equal a non-square number.

How can we prove that there is no solution to this equation?

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.

Can we find a solution to this equation by using complex numbers?

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.

Are there any other ways to prove that this equation has no solution?

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.

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