Question Number 23050 by jazary last updated on 25/Oct/17
![how can demonstred that ∀a,b,c∈N a^2 +b^2 =c^2 ⇒ abc≡0[60]](https://www.tinkutara.com/question/Q23050.png)
$${how}\:{can}\:{demonstred}\:{that}\: \\ $$$$\:\:\:\forall{a},{b},{c}\in\mathbb{N}\: \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={c}^{\mathrm{2}} \:\:\Rightarrow \\ $$$${abc}\equiv\mathrm{0}\left[\mathrm{60}\right]\:\: \\ $$
Answered by Rasheed.Sindhi last updated on 29/Oct/17
![abc≡0[60]⇒60∣abc ⇒(4.3.5)∣abc ⇒4∣abc ∧ 3∣abc ∧ 5∣abc ^• 4∣abc⇒4∣ab ∨ 4∣bc ∨ 4∣ ac......(i) ^• 3∣abc⇒3∣a ∨ 3∣b ∨ 3∣c...........(ii) ^• 5∣abc⇒5∣a ∨ 5∣b ∨ 5∣c..........(iii) We have to prove: Case-1:(i):4∤ab ∧ 4∤bc⇒4∣ac This can be proved like Case-2 Case-2:(ii):3∤a ∧ 3∤b⇒3∣c 3∤a ∧ 3∤b means a and b or of 3k+1 or 3k+2 type ⇒a^2 and b^2 are of 3k+1 type c=(√(a^2 +b^2 )) =(√((3k+1)+(3l+1))) =(√(3m+2)) But 3m+2 is not perfect square ∴ If c∈N⇒a^2 +b^2 is perfect square ⇒a^2 and b^2 are not both of type 3k+1. ∴ a or b aren′t of type 3k+1 or 3k+2. ∴ a or b is 3k-type ∴ 3∣a or 3∣b ∴3∣abc Case-3:(iii):5∤a ∧ 5∤b⇒5∣c This case can also be proved like Case-2 4∣abc Case-1 3∣abc Case-2 5∣abc Case-3 Hence 60∣abc Or abc≡0[60]](https://www.tinkutara.com/question/Q23360.png)
$${abc}\equiv\mathrm{0}\left[\mathrm{60}\right]\Rightarrow\mathrm{60}\mid{abc}\:\: \\ $$$$\Rightarrow\left(\mathrm{4}.\mathrm{3}.\mathrm{5}\right)\mid{abc} \\ $$$$\Rightarrow\mathrm{4}\mid{abc}\:\wedge\:\mathrm{3}\mid{abc}\:\wedge\:\mathrm{5}\mid{abc} \\ $$$$\:^{\bullet} \mathrm{4}\mid{abc}\Rightarrow\mathrm{4}\mid{ab}\:\vee\:\mathrm{4}\mid{bc}\:\vee\:\mathrm{4}\mid\:{ac}……\left(\mathrm{i}\right) \\ $$$$\:^{\bullet} \mathrm{3}\mid{abc}\Rightarrow\mathrm{3}\mid{a}\:\vee\:\mathrm{3}\mid{b}\:\vee\:\mathrm{3}\mid{c}………..\left(\mathrm{ii}\right) \\ $$$$\:^{\bullet} \mathrm{5}\mid{abc}\Rightarrow\mathrm{5}\mid{a}\:\vee\:\mathrm{5}\mid{b}\:\vee\:\mathrm{5}\mid{c}……….\left(\mathrm{iii}\right) \\ $$$$\mathrm{We}\:\mathrm{have}\:\mathrm{to}\:\mathrm{prove}: \\ $$$$\mathrm{Case}-\mathrm{1}:\left(\mathrm{i}\right):\mathrm{4}\nmid{ab}\:\wedge\:\mathrm{4}\nmid{bc}\Rightarrow\mathrm{4}\mid{ac} \\ $$$$\:\:\:\mathrm{This}\:\mathrm{can}\:\mathrm{be}\:\mathrm{proved}\:\mathrm{like}\:\mathrm{Case}-\mathrm{2} \\ $$$$\mathrm{Case}-\mathrm{2}:\left(\mathrm{ii}\right):\mathrm{3}\nmid{a}\:\wedge\:\mathrm{3}\nmid{b}\Rightarrow\mathrm{3}\mid{c} \\ $$$$\mathrm{3}\nmid{a}\:\wedge\:\mathrm{3}\nmid{b}\:\mathrm{means}\:{a}\:\mathrm{and}\:{b}\:\mathrm{or}\:\mathrm{of} \\ $$$$\mathrm{3k}+\mathrm{1}\:\mathrm{or}\:\:\mathrm{3k}+\mathrm{2}\:\mathrm{type} \\ $$$$\Rightarrow{a}^{\mathrm{2}} \:\mathrm{and}\:{b}^{\mathrm{2}} \:\mathrm{are}\:\mathrm{of}\:\mathrm{3k}+\mathrm{1}\:\mathrm{type} \\ $$$$\mathrm{c}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:}\:\:=\sqrt{\left(\mathrm{3k}+\mathrm{1}\right)+\left(\mathrm{3}{l}+\mathrm{1}\right)} \\ $$$$\:\:\:=\sqrt{\mathrm{3m}+\mathrm{2}} \\ $$$$\mathrm{But}\:\mathrm{3m}+\mathrm{2}\:\mathrm{is}\:\mathrm{not}\:\mathrm{perfect}\:\mathrm{square} \\ $$$$\therefore\:\mathrm{If}\:\:\mathrm{c}\in\mathbb{N}\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:\mathrm{is}\:\mathrm{perfect}\:\mathrm{square} \\ $$$$\Rightarrow{a}^{\mathrm{2}} \mathrm{and}\:{b}^{\mathrm{2}} \:\mathrm{are}\:\mathrm{not}\:\mathrm{both}\:\mathrm{of}\:\mathrm{type} \\ $$$$\:\:\:\:\:\mathrm{3k}+\mathrm{1}. \\ $$$$\therefore\:{a}\:\mathrm{or}\:{b}\:\mathrm{aren}'\mathrm{t}\:\mathrm{of}\:\mathrm{type}\:\mathrm{3k}+\mathrm{1}\:\mathrm{or} \\ $$$$\mathrm{3k}+\mathrm{2}. \\ $$$$\therefore\:{a}\:\mathrm{or}\:{b}\:\mathrm{is}\:\mathrm{3k}-\mathrm{type} \\ $$$$\therefore\:\mathrm{3}\mid{a}\:\mathrm{or}\:\mathrm{3}\mid{b} \\ $$$$\therefore\mathrm{3}\mid{abc} \\ $$$$\mathrm{Case}-\mathrm{3}:\left(\mathrm{iii}\right):\mathrm{5}\nmid{a}\:\wedge\:\mathrm{5}\nmid{b}\Rightarrow\mathrm{5}\mid{c} \\ $$$$\:\:\mathrm{This}\:\mathrm{case}\:\mathrm{can}\:\mathrm{also}\:\mathrm{be}\:\mathrm{proved}\:\mathrm{like}\:\mathrm{Case}-\mathrm{2} \\ $$$$\mathrm{4}\mid{abc}\:\:\:\mathrm{Case}-\mathrm{1} \\ $$$$\mathrm{3}\mid{abc}\:\:\:\mathrm{Case}-\mathrm{2} \\ $$$$\mathrm{5}\mid{abc}\:\:\:\:\mathrm{Case}-\mathrm{3} \\ $$$$\mathrm{Hence}\:\mathrm{60}\mid{abc} \\ $$$$\mathrm{Or}\:{abc}\equiv\mathrm{0}\left[\mathrm{60}\right] \\ $$