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Using congruences, I seek to prove two things:

1) $x^2 - 4y^2 = 3$ has no solutions in integers $x,y,z$.

I think this can be done using modulo 4? How so?

2) $3x^3 - 7y^3 + 21 z^3 = 2$ has no solutions in integers $x,y,z$.

Not sure how to attack this one...

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3 Answers 3

up vote 3 down vote accepted

Your first equation is $x^2 - 4y^2 =3$. Mod 4, this is asking for solutions to $x^2 =3$ mod $4$. Checking with $x= 0,1,2,3$ shows that there are no solutions because $0^2 = 0 , 1^2 = 1,2^2 =0, 3^2 = 1$.

For the second one if you work mod 7, then the equation becomes $3x^3 = 2$. Trying this with $x = 0, \ldots 6$ shows that there are no solutions so you are done.

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(1) If $x$ is even, is $x^2-4y^2$ odd or even? If $x$ is odd, can $x^2-3$ be divisible by $4$ if $x$ is an integer?

(2) $3x^3-7y^3+21z^3=7(3z^3-y^3)+3x^3$; can $3x^3-2$ be divisible by $7$ if $x$ is an integer?

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Hint $\rm\ (1)\ \ mod\ 4\!:\ odd^2 \equiv (\pm1)^2\equiv 1,\ even^2\equiv 0,\:$ so $\rm\:x^2\not\equiv 3.$

and $\rm\ \ \ \:\! (2)\ \ mod\ 7\!:\ x\not\equiv 0\: \Rightarrow\: x^6\equiv 1\: \Rightarrow\: x^3\equiv \pm1\ $ by $\mu\!$ Fermat; directly $\rm\{\pm1,\pm2,\pm3\}^3 \!\equiv \pm 1$

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