Question Number 70525 by Rasheed.Sindhi last updated on 05/Oct/19
$${If}\:{x},{y},{z}\:{is}\:{a}\:{primitive}\:{Pythagorean} \\ $$$${triple},{prove}\:{that}\:{x}+{y}\:{and}\:{x}−{y}\:{are} \\ $$$${congruent}\:{modulo}\:\mathrm{8}\:{to}\:{either}\:\mathrm{1}\:{or}\:\mathrm{7}. \\ $$
Answered by mind is power last updated on 07/Oct/19
![x=2ab y=a^2 −b^2 z=a^2 +b^2 x,y,z primitive ⇒gcd(a,b)=1 and either a= 2k &b=2w+1 because if gcd(a,b)=a≥2 ⇒a∣x,y,z abvious and if a=b[2]⇒2∣x,y,z abvious x=2(2k)(2w+1)=8kw+4k=4k(8) y=(4k^2 )−4w^2 −4w−1 y=4k^2 −4w(w+1)−1=4k^2 −1(8) cause 2∣w(w+1),∀w∈IN just use either w=2s or w=2s+1 to see that if kor k=0(2)⇒x=0(8),y=−1(8)⇒ { ((x−y=1(8))),((x+y=−1(8)=7(8))) :} k=1(2)⇒x=4(8),y=3(8)⇒ { ((x−y=1(8))),((x+y=7(8))) :} ⇒x+y ,x−y congurent modulo 8 to either 1 or 7](https://www.tinkutara.com/question/Q70708.png)
$${x}=\mathrm{2}{ab} \\ $$$${y}={a}^{\mathrm{2}} −{b}^{\mathrm{2}} \\ $$$${z}={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$${x},{y},{z}\:{primitive}\:\Rightarrow{gcd}\left({a},{b}\right)=\mathrm{1}\:{and}\:\:{either}\:{a}=\:\mathrm{2}{k}\:\&{b}=\mathrm{2}{w}+\mathrm{1}\: \\ $$$${because}\:{if}\:{gcd}\left({a},{b}\right)={a}\geqslant\mathrm{2} \\ $$$$\Rightarrow{a}\mid{x},{y},{z}\:{abvious} \\ $$$${and}\:{if}\:{a}={b}\left[\mathrm{2}\right]\Rightarrow\mathrm{2}\mid{x},{y},{z}\:{abvious} \\ $$$${x}=\mathrm{2}\left(\mathrm{2}{k}\right)\left(\mathrm{2}{w}+\mathrm{1}\right)=\mathrm{8}{kw}+\mathrm{4}{k}=\mathrm{4}{k}\left(\mathrm{8}\right) \\ $$$${y}=\left(\mathrm{4}{k}^{\mathrm{2}} \right)−\mathrm{4}{w}^{\mathrm{2}} −\mathrm{4}{w}−\mathrm{1} \\ $$$${y}=\mathrm{4}{k}^{\mathrm{2}} −\mathrm{4}{w}\left({w}+\mathrm{1}\right)−\mathrm{1}=\mathrm{4}{k}^{\mathrm{2}} −\mathrm{1}\left(\mathrm{8}\right) \\ $$$${cause}\:\mathrm{2}\mid{w}\left({w}+\mathrm{1}\right),\forall{w}\in{IN}\:{just}\:{use}\:{either}\:{w}=\mathrm{2}{s}\:{or}\:{w}=\mathrm{2}{s}+\mathrm{1}\:{to}\:{see}\:{that} \\ $$$$\boldsymbol{{if}}\:\boldsymbol{{ko}}{r}\:{k}=\mathrm{0}\left(\mathrm{2}\right)\Rightarrow{x}=\mathrm{0}\left(\mathrm{8}\right),{y}=−\mathrm{1}\left(\mathrm{8}\right)\Rightarrow\begin{cases}{{x}−{y}=\mathrm{1}\left(\mathrm{8}\right)}\\{{x}+{y}=−\mathrm{1}\left(\mathrm{8}\right)=\mathrm{7}\left(\mathrm{8}\right)}\end{cases} \\ $$$$ \\ $$$${k}=\mathrm{1}\left(\mathrm{2}\right)\Rightarrow{x}=\mathrm{4}\left(\mathrm{8}\right),{y}=\mathrm{3}\left(\mathrm{8}\right)\Rightarrow\begin{cases}{{x}−{y}=\mathrm{1}\left(\mathrm{8}\right)}\\{{x}+{y}=\mathrm{7}\left(\mathrm{8}\right)}\end{cases} \\ $$$$\Rightarrow{x}+{y}\:,{x}−{y}\:{congurent}\:{modulo}\:\mathrm{8}\:{to}\:{either}\:\mathrm{1}\:{or}\:\mathrm{7} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Rasheed.Sindhi last updated on 08/Oct/19
$$\mathcal{N}{o}\:{doubt},\:{mind}\:{is}\:{power}! \\ $$