If $\gcd(a,b)=1$, how can I find the values that $\gcd(a+b,a^2+b^2)$ can possibly take? I can't find a way to use any of the elemental divisibility and gcd theorems to find them.
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2I added an answer which proves that $ \rm: (a!+!b,\ a^2!+!b^2),=,(a!+!b,,2(a,b)^2!):$ for any value of $\rm:(a,b).:$ Therefore $ \rm: (a!+!b,\ a^2!+!b^2),=,(a!+!b,,2):$ in your special case when $\rm:(a,b) = 1.:$ – Math Gems Feb 21 '13 at 19:28
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We have $a^2 + b^2 - a(a+b) = b^2 - ab = -b (a-b)$ and $a^2 + b^2 - b(a+b) = a^2 - ba = a (a-b)$.
So if $d$ divides both $a+b$ and $a^2+b^2$, then $d$ divides $$\gcd(a (a-b), b (a-b)) = \gcd(a, b) (a-b) = a - b.$$
So $d$ divides $a+b + a - b = 2a$ and $a+b - (a - b) = 2b$.
So $d$ divides $2\gcd(a,b)=2$.
So the possibilities for the $\gcd$ appear to be $1$ and $2$, and both clearly occur.
Andreas Caranti
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3A well-earned +1, too - this is much cleaner than the approach I was taking. – Steven Stadnicki Feb 18 '13 at 23:26
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$$\begin{align} \gcd(a+b, a^2 + b^2) &= \gcd(a+b, a^2 + b^2 - a(a+b)) \\&= \gcd(a+b, b^2 - ab) \\&= \gcd(a+b, b^2 - ab + b(a+b)) \\&= \gcd(a+b, 2b^2) \end{align} $$
Now, $\gcd(a+b,b) = \gcd(a,b) = 1$, so we can get rid of the factors of $b$ and have
$$ \gcd(a+b, a^2 + b^2) = \gcd(a+b, 2) $$
The strategy I used was still the basic idea of the Euclidean algorithm; since I couldn't compare numeric values, I instead simplified by working to eliminate the variable $a$, starting with the largest power of $a$.
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Which property or theorem do you use to get rid of the factors of b? – Richard Codwater Feb 19 '13 at 02:23
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There are a few things you can do with products. In this case, I invoked: $$\gcd(x,y)=1 \implies \gcd(x,yz) = \gcd(x,z)$$ – Feb 19 '13 at 03:16
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2@Rob In fact one can give an exact formula $\rm: (a!+!b,\ a^2!+!b^2),=,(a!+!b,,2(a,b)^2!):$ even for the case $\rm:(a,b)\ne 1,: $ see my answer. – Math Gems Feb 21 '13 at 19:14
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Since $\gcd(a,b)=1$, Bezout's Identity says we have an $x$ and $y$ so that $$ ax+by=1\tag{1} $$ Note that $$ \begin{align} 2a^2&=(a^2+b^2)+(a+b)(a-b)\\ 2ab&=(a+b)^2-(a^2+b^2)\\ 2b^2&=(a^2+b^2)-(a+b)(a-b)\\ \end{align}\tag{2} $$ Therefore, incorporating $(1)$ and $(2)$, $$ \begin{align} 2 &=2(ax+by)^2\\ &=2a^2x^2+4abxy+2b^2y^2\\ &=\Big((a^2+b^2)+(a+b)(a-b)\Big)x^2\\ &+2\Big((a+b)^2-(a^2+b^2)\Big)xy\\ &+\Big((a^2+b^2)-(a+b)(a-b)\Big)y^2\\ &=\color{#00A000}{(x-y)^2}\color{#C00000}{(a^2+b^2)} +\color{#00A000}{((x^2-y^2)(a-b)+2xy(a+b))}\color{#C00000}{(a+b)}\tag{3} \end{align} $$ Equation $(3)$ says that $$ \gcd(a+b,a^2+b^2)\,|\,2\tag{4} $$ Note that $\gcd(1+2,1^2+2^2)=1$ and $\gcd(1+3,1^2+3^2)=2$, so both $1$ and $2$ are possible.
Martin Sleziak
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robjohn
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One can also prove it more simply without Bezout, using only gcd laws and Euclid's lemma: $$\begin{array}{l}\rm d,|\ ,\color{#C00}{a+b}\ \rm d,|,a^2!!+b^2\ \end{array}\ \Rightarrow\ \ \rm\begin{array}{l} d,|,2a^2, = , \color{#C00}{a^2!!-!b^2}, +a^2!!+b^2\ \rm d,|,2b^2, = , \color{#C00}{b^2!!-!a^2},+a^2!!+b^2\end{array}\ \Rightarrow\ \ \rm d,|,(2a^2,2b^2)=2(a^2,b^2) = 2$$ – Math Gems Feb 21 '13 at 19:12
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@MathGems: Indeed, $(2)$ above is the first $\Rightarrow$, and $(3)$ essentially reproves Euclid's Lemma. I figured there would be other non-Bezout proofs, and so the challenge was to find the linear combination of $a+b$ and $a^2+b^2$ that equals $2$. :-) – robjohn Feb 21 '13 at 20:27
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Put $\,\rm (a,b)=1\,$ below.
Theorem $\rm\,\ \ (a\!+\!b,\ a^2\!+\!b^2)\, =\, (\color{#c00}{2a^2,\ \ 2ab,\ \ 2b^2},\ a\!+\!b)\, \overset{\rm\color{#c00}E}=\, (\color{#c50}{2(a,b)^2}\!,\ a\!+\!b)$
Proof $\rm\ mod\ a\!+\!b\!:\ a^2\!+\!b^2 \equiv \color{#c00}{2a^2\! \equiv -2ab \equiv 2b^2}\ $ by $\rm\,a\!+\!b\,$ divides $\rm\color{#0A0}{green}$ terms below
$$\quad\ \ \ \rm a^2\!+\!b^2 = (\color{#0A0}{b^2\!-\!a^2})+\color{#c00}{2a^2} = (\color{#0A0}{a\!+\!b})^2\!\color{#c00}{-2ab} = (\color{#0A0}{a^2\!-\!b^2})+\color{#c00}{2b^2} $$
Bill Dubuque
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Math Gems
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We used $\rm: (c,d_i) = (c,d)\ $ if $\rm\ d_i\equiv d\ (mod\ c),,$ cf. here, e.g. $\rm: (c,d) = (c,, d\ mod\ c),:$ the recursive step (descent) at the heart of the Euclidean algorithm (essentially, what we used). The final equality $\rm\ \color{#c50}{2(a,b)^2}\color{#c00}{ \overset{\rm\color{#c00}E}=, 2(a^2,ab,b^2)}:$ is by gcd laws (associative,distributive, etc), i.e.
$$\rm (a,b)(a,b) = ((a,b)a,(a,b)b) = (a^2,ba,ab,b^2) = (a^2,ab,b^2)\ \ \ \ $$
– Bill Dubuque May 29 '22 at 05:20
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$$ \gcd(a+b,a^2+b^2) \mid \gcd((a+b)(a-b), a^2+b^2) = \gcd(a^2-b^2, a^2+b^2) \mid \gcd [ ( a^2+b^2)+ (a^2-b^2) , ( a^2+b^2)+ (a^2-b^2) ]=2 \gcd(a^2,b^2)=2$$
Now it is easy to check that both 1 and 2 are possible...
Martin Sleziak
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N. S.
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That's essentially the same method as in my un-Bezouted comment to Rob's answer. – Math Gems Feb 21 '13 at 19:30
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Note that $a^2+b^2=(a+b)^2-2ab$. Thus $a^2+b^2\equiv -2ab\mod a+b$ and you need to find $(a+b,-2ab)=(a+b,2ab)$. Can you move on?
ADD Note that $a,b$ cannot be both even. If one is odd and the other is even, then $2\not\mid a+b$, so $(a+b,2ab)=(a+b,ab)$. But if $p>2$ is a prime with $p\mid a+b,ab$ then $p\mid a$ or $p\mid b$, since $p\mid ab$. But in any case, since $p\mid a+b$, this would give $p \mid b$ (if $p\mid a$) or $p \mid a$ (if $p\mid b)$, contrary to $(a,b)=1$. Thus $(a+b,ab)=1$.
If $a,b$ are both odd, then $a+b$ is even and $2\mid (a+b,2ab)$. And again, if $p$ is an odd prime factor $p\mid 2ab\implies p\mid ab$ and the same argument above goes through. Thus whenever $(a,b)=1$, $$(a+b,a^2+b^2)=\begin{cases} 1 &\text{if one of } a,b \text{ is odd}\\2&\text{if both} a,b\text{ are odd}\end{cases}$$
Pedro
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