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%(NZM, Problem ) 

\ctln{\bf Math 445 Homework 5}

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\ctln{Due Friday, October 10}

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\item{17.} Show that if $a^2+2b^2=c^2$ and $p|(a,c)$ for some prime $p$, then $p|b$.

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\item{} (Note: $p=2$ is treated differently...)


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\item{} If $p|(a,c)$, then $a=px$, $c=py$ for some integers $x,y$, so $a^2+2b^2=c^2$ implies $2b^2=c^2-a^2=(p^2)(y^2-x^2)$, so 
$p^2|2b^2$. If $p=2$, this in turn implies that, since $4|2b^2$, $2|b^2$ so, since $2$ is prime, $2|b$ or $2|b$, so $2|b$.
If $p\neq 2$, then $p^2|2b^2$ implies $p|2b^2$, and since $(p,2)=1$, $p|b^2$, so $p|b$ of $p|b$, so $p|b$.

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\item{} So in all cases, $p$ prime and $p|(a,c)$ implies $p|b$.

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\item{18.} Show that if $a^2+2b^2=c^2$ and $(a,c)=1$, then $a$ and $c$ are odd, and $b$ is even.

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\item{} If $a$ is even, $a=2x$, then $c^2=a^2+2b^2=4x^2+2b^2=2(2x^2+b^2)$ is even, so $c$ is even, so $2|(a,c)$ a contradiction. 
If $c$ is even, $c=2y$, then
$a^2=c^2-2b^2=4y^2-2b^2=2(2y^2-b^2$ is even, so $a$ is even, which is again a contradiction.

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\item{} So both $a$ and $c$ are odd, $a=2k+1$, $c=2m+1$.
If $b$ is odd, $b=2n+1$, then we have 

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\item{} $0=a^2+2b^2-c^2=(2k+1)^2+2(2m+1)^2-(2n+1)^2$

\hfill $=4(k^2+k+2m^2+2m-n^2-n)+2$

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\item{} so $0\equiv 2$ (mod $4$), which is absurd. So $b$ must be even.

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\item{19.} Continuing the analysis: with $(a,c)=1$, writing 

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\ctln{$2b^2=c^2-a^2=(c-a)(c+a)=(2u)(2v)$ ,}

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\item{} show that $(u,v)=1$, one of $u,v$ is odd and the other even, and 
$u=2x^2,v=y^2$ or $u=y^2,v=2x^2$ for some integers $x,y$.

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\item{} Since both $a$ and $c$ are odd, $c-a=2u$ and $c+a=2v$ are both even. Since (adding these two equations)
$2c=2u+2v$, we have $c=u+v$, and since (subtracting the two equations) $2a=2v-2u$, we have $a=v-u$.
If $d|u,v$, then $d|u+v,v-u$, so $d|a,c$ so $d|(a,c)=1$; so the largest integer dividing both
$u$ and $v$ is $1$, so $(u,v)=1$. 

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\item{} Since $b$ is even, $b=2z$ for some integer $z$, and then 

\ctln{$8z^2=2b^2=c^2-a^2=(c-a)(c+a)=(2u)(2v)=4uv$, }

\item{} so $2z^2=uv$, so $2|uv$, so $2|u$ or $2|v$; for the sake of concreteness, as assume $2|u$
(the other case is similar). Note that we then cannot have $2|v$, since then $2|(u,v)=1$.
So if $u$ is even, then $v$ is odd (the opposite situation is similar). So $u=2w$, and
$(u,v)=1$ implies $(w,v)=1$; anything which divides $w$ and $v$ also divides $u$ and $v$.
Then $2z^2=uv=2wv$ implies $z^2=wv$. Since $(w,v)=1$, our result from class implies that
$w=x^2$, $v=y^2$ for some integers $x,y$, and so $u=2x^2$, $v=y^2$, as desired.

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\item{} (The opposite case, $u$ odd and $v$ even, yields the other alternative $u=y^2$, $v=2x^2$.)

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\item{20.} Finishing the analysis: show that $(2x^2-y^2)^2+2(2xy)^2=(2x^2+y^2)^2$, and every solution
to $a^2+2b^2=c^2$ with $a,b,c\in{\Bbb N}$ has this form: 

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\ctln{$a=|2x^2-y^2|$, $b=2xy$, $c=2x^2+y^2$, with $x,y\geq 0$ and $y$ odd .}

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\item{} For this problem, the assumption $(a,c)=1$ was intended to continue....

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\item{} We have that if $a^2+2b^2=c^2$ with $(a,c)=1$, then 

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$a=v-u=y^2-2x^2$, $b^2=2uv=4x^2y^2=(2xy)^2$, so

\hfill $b=2xy$, and $c=v+u=y^2+2x^2$

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\item{} (in the case $u$ even and $v$ odd, \underbar{or}

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$a=v-u=2x^2-y^2$, $b^2=2uv=4x^2y^2=(2xy)^2$, so

\hfill $b=2xy$, and $c=v+u=2x^2+y^2$

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\item{} (in the case $u$ odd and $v$ even.

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\item{} We can unify these two cases by writing $a=|2x^2-y^2|$, $b=2xy$, and $c=2x^2+y^2$ for integers $x,y$.
So if $a^2+2b^2=c^2$ with $(a,c)=1$, then $a=|2x^2-y^2|$, $b=2xy$, and $c=2x^2+y^2$ for integers $x,y$.

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\item{} Note that for any integers $x,y$, these numbers $a,b,c$ \underbar{do} satisfy $a^2+2b^2=c^2$:

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\item{} $|2x^2-y^2|^2+2(2xy)^2=4x^4-4x^2y^2+y^4+8x^2y^2=4x^4+4x^2y^2+y^4$

\hfill $=(2x^2)^2+2(2x^2)(y^2)+(y^2)^2=(2x^2+y^2)^2$ .

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\item{} The last thing to note is that $y$ must be odd; if $y=2z$, then $2|2(x^2-2z^2)=2x^2-y^2=\pm a$ and 
$2|2(x^2+2z^2)=2x^2+y^2=c$, so $2|(a,c)=1$, a contradiction.


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