Prove :x²+y²=1992 ,no solution

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SUMMARY

The equation $x^2 + y^2 = 1992$ has been proven to have no solutions in natural numbers ($x, y \in \mathbb{N}$). This conclusion is derived from the properties of sums of squares and the analysis of the number 1992 in relation to its prime factorization. Specifically, 1992 can be expressed as $2^3 \times 3 \times 83$, and since it contains a prime of the form $4k + 3$ raised to an odd power, it cannot be represented as a sum of two squares.

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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
 
Last edited:

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