1
$\begingroup$

Working independently on some practice problems in preparation for my class exam and got stuck on the following question:

Show there are no solutions in integers $x$, $y$, and $z$ to the diophantine equation $x^2 + 2y^2 = 8z + 5.$

I know that we have to prove it using modular arithmetic but I'm not sure how to go about solving this problem. Could anyone help? Thanks!

$\endgroup$

2 Answers 2

Reset to default
8
$\begingroup$

If you look at the equation mod $8$, note that squares are either $0,1$ or $4$ mod 8. Thus the left hand side can only attain the values $0,1,2,3,4$ or $6$ mod $8$ and hence will never be equal to the right hand side, which is $5$ mod $8$.

$\endgroup$
3
  • $\begingroup$ $x^2+2y^2$ can't be $6\bmod 8$. $\endgroup$
    – Bernard
    May 29, 2019 at 22:19
  • $\begingroup$ @Bernard $2^2+2 \times 1^2 = 6$ $\endgroup$
    – Henry
    May 29, 2019 at 22:25
  • $\begingroup$ @Henry: you're right. For my excuse, it's getting late here… $\endgroup$
    – Bernard
    May 29, 2019 at 22:26
1
$\begingroup$

If you have $x^2+2y^2=8z+5$, by looking $\bmod 2$, you have that $x$ has to be odd.

Now (a very useful fact for me) we have that $odd^2\equiv 1\mod 8$. In particular $x^2\equiv 1\mod 8$.

So, if you look your original equation $\bmod 8$ you get $$1+2y^2=5\bmod 8$$ This implies that $2y^2\equiv 4\mod 8$ and by dividing by $2$, you get $y^2\equiv 2\mod 4$. The last thing is imposible, so there are no solutions.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .