Prove that the equation $x^2 + y^2 + z^2 = x + y + z + 1$ has no rational solutions.
Problem
Source: Mediterranean MO 2003
Tags: algebra proposed, algebra
dgreenb801
31.10.2010 16:17
It is equivalent to $(2x-1)^2+(2y-1)^2+(2z-1)^2=7$. But if this had a solution then so would
$a^2+b^2+c^2=7d^2$ for some positive integers $a$,$b$,$c$, and $d$. In fact, we can assume it is a reduced solution, that is, $a$,$b$,$c$, and $d$ do not all have a common factor. Now squares are either $0$,$1$,$4$, or $9$ mod $16$. Using this we get there are no reduced solutions and thus no solutions at all.
Golub_Srecko
17.05.2017 22:03
dgreenb801 wrote:
It is equivalent to $(2x-1)^2+(2y-1)^2+(2z-1)^2=7$. But if this had a solution then so would
$a^2+b^2+c^2=7d^2$ for some positive integers $a$,$b$,$c$, and $d$. In fact, we can assume it is a reduced solution, that is, $a$,$b$,$c$, and $d$ do not all have a common factor. Now squares are either $0$,$1$,$4$, or $9$ mod $16$. Using this we get there are no reduced solutions and thus no solutions at all.
Integers? If numbers were integers mod2 would work right away... Unfortunately we have to prove equation has no rational solutions.
Garfield
17.05.2017 22:31
Golub_Srecko wrote: dgreenb801 wrote:
It is equivalent to $(2x-1)^2+(2y-1)^2+(2z-1)^2=7$. But if this had a solution then so would
$a^2+b^2+c^2=7d^2$ for some positive integers $a$,$b$,$c$, and $d$. In fact, we can assume it is a reduced solution, that is, $a$,$b$,$c$, and $d$ do not all have a common factor. Now squares are either $0$,$1$,$4$, or $9$ mod $16$. Using this we get there are no reduced solutions and thus no solutions at all.
Integers? If numbers were integers mod2 would work right away... Unfortunately we have to prove equation has no rational solutions. Sketch is ok ,he didn't wrote step $2x-1=\frac{p}{q}$,$2y-1=\frac{p_{1}}{q_{1}}$,$2z-1=\frac{p_{2}}{q_{2}}$ so put it on same denominator and get $\frac{(p\cdot q_{1}\cdot q_{2})^{2}+(p_{1}\cdot q\cdot q_{2})^{2}+(q\cdot q_{1}\cdot p_{2})^{2}}{(q\cdot q_{1}\cdot q_{2})^{2}}=7$ so $(p\cdot q_{1}\cdot q_{2})^{2}+(p_{1}\cdot q\cdot q_{2})^{2}+(q\cdot q_{1}\cdot p_{2})^{2}=7\cdot (q\cdot q_{1}\cdot q_{2})^{2}$