MHB Prove :x²+y²=1992 ,no solution

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The equation \(x^2 + y^2 = 1992\) has no solutions in natural numbers \(x\) and \(y\). This is demonstrated by analyzing the properties of numbers expressible as sums of two squares. Specifically, a number can be expressed as a sum of two squares if it has no prime factors of the form \(4k + 3\) raised to an odd power. Since 1992 includes the prime factorization \(2^3 \times 3^1 \times 83^1\), the presence of the factor \(3^1\) indicates it cannot be expressed as such. Therefore, the conclusion is that there are no natural number solutions for the equation.
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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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Thread 'Area of Overlapping Squares'

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