Find prime numbers $y$ and $z$ if $x$ is an integer and $x^2-y^2-z^2=2017.$
$x^2-y^2-z^2$ is not factorable, so what should I do? My mind is blank here. Any help is appreciated.
Find all prime numbers $y$ and $z$ if $x$ is an integer and $x^2-y^2-z^2=2017.$
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algebra-precalculus
prime-numbers
square-numbers
sums-of-squares
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0Are you looking for all solutions or just one? Hint: write it as $(x+y)(x-y)=2017+z^2$ and look for suitable divisors of the right hand. First example I tried worked...but I expect there are many other examples. – 2017-01-19
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0@lulu All, sorry it wasn't in the problem. – 2017-01-19
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0@MonsieurGalois Yes. – 2017-01-19
1 Answers
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We have that $$a^2\equiv 1\lor 0\pmod{4}$$ We also have that $$2017\equiv 1\pmod{4}$$.Now try to choose $x^2,y^2,z^2$ so that you get $1\pmod{4}$.You can see that the only solution is if $y^2\equiv 0\pmod{4}$ and $z^2\equiv 0\pmod{4}$ but since $z,y$ are prime we have that $z=y=2$ so the only solution is $x^2-8=2017,x^2=2025,x=45$
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0Do you know if there could be an algebraic way to prove the answer? – 2017-01-19
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0@ReginaDea Doesn't seem probable,usually when you have prime numbers you use divisibility/modular arithmetic. – 2017-01-19