The number of solutions of the equation x^2 + | vedclass

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(A) Consider the equation $x^2 + y^2 = a^2 + b^2 + c^2$ where $x, y, a, b, c$ are primes.Case $1$: If all primes are odd,then $x^2, y^2, a^2, b^2, c^2

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$0$

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(A) Consider the equation $x^2 + y^2 = a^2 + b^2 + c^2$ where $x, y, a, b, c$ are primes.Case $1$: If all primes are odd,then $x^2, y^2, a^2, b^2, c^2 \equiv 1 \pmod{4}$ or $1 \pmod{8}$.Specifically,for any odd prime $p$,$p^2 \equiv 1 \pmod{8}$.Then $x^2 + y^2 \equiv 1 + 1 = 2 \pmod{8}$ and $a^2 + b^2 + c^2 \equiv 1 + 1 + 1 = 3 \pmod{8}$.Since $2 
ot\equiv 3 \pmod{8}$,there are no solutions where all variables are odd primes.Case $2$: If at least one variable is $2$.If $x=2$,then $4 + y^2 = a^2 + b^2 + c^2$. If $y$ is also $2$,then $8 = a^2 + b^2 + c^2$. The only primes whose squares are less than $8$ are $2$. If $a=b=c=2$,then $a^2 + b^2 + c^2 = 4 + 4 + 4 = 12 
eq 8$. If any are $2$,we cannot satisfy the sum.Testing small primes,we find no combination of prime numbers satisfies the equation.Thus,the number of solutions is $0$.

The number of solutions of the equation $x^2 + y^2 = a^2 + b^2 + c^2$,where $x, y, a, b, c$ are all prime numbers,is

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