Menu Close

if-a-b-c-Z-and-a-2-b-2-c-2-then-3-ab-

Tinku Tara 0 Comments
Pin




Question Number 219289 by Nicholas666 last updated on 22/Apr/25
if a,b,c ∈ Z , and a^2 + b^2 = c^2 , then 3∣(ab) = ?
$$ \\ $$$$\:\:\:\:{if}\:\:\:\:\:\:\:\:\:\:\:{a},{b},{c}\:\in\:\mathbb{Z}\:\:\:, \\ $$$$\:\:\:\:\:{and}\:\:\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:=\:\:{c}^{\mathrm{2}} \:\:\:, \\ $$$$\:\:\:\:\:\:\:\:{then}\:\:\:\:\:\:\mathrm{3}\mid\left({ab}\right)\:=\:? \\ $$$$ \\ $$
Commented by Nicholas666 last updated on 22/Apr/25
thanks
$${thanks} \\ $$
Answered by A5T last updated on 22/Apr/25
Any perfect square is equivalent to 0 or 1 (mod 3) ⇒a^2 +b^2 ≡ 0+0 ,0+1, 1+0 ,1+1 (mod 3) But a^2 +b^2 is also a perfect square, hence cannot be equivalent to 1+1=2(mod 3). ⇒ (a,b)≡(0,0);(0,1);(1,0) (mod 3) ⇒At least one of a,b is divisible by 3. Hence, 3∣(ab)
$$\mathrm{Any}\:\mathrm{perfect}\:\mathrm{square}\:\mathrm{is}\:\mathrm{equivalent}\:\mathrm{to}\:\mathrm{0}\:\mathrm{or}\:\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{3}\right) \\ $$$$\Rightarrow\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \:\equiv\:\mathrm{0}+\mathrm{0}\:,\mathrm{0}+\mathrm{1},\:\mathrm{1}+\mathrm{0}\:,\mathrm{1}+\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{3}\right) \\ $$$$\mathrm{But}\:\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \:\mathrm{is}\:\mathrm{also}\:\mathrm{a}\:\mathrm{perfect}\:\mathrm{square},\:\mathrm{hence}\:\mathrm{cannot} \\ $$$$\mathrm{be}\:\mathrm{equivalent}\:\mathrm{to}\:\mathrm{1}+\mathrm{1}=\mathrm{2}\left(\mathrm{mod}\:\mathrm{3}\right). \\ $$$$\Rightarrow\:\left(\mathrm{a},\mathrm{b}\right)\equiv\left(\mathrm{0},\mathrm{0}\right);\left(\mathrm{0},\mathrm{1}\right);\left(\mathrm{1},\mathrm{0}\right)\:\left(\mathrm{mod}\:\mathrm{3}\right) \\ $$$$\Rightarrow\mathrm{At}\:\mathrm{least}\:\mathrm{one}\:\mathrm{of}\:\mathrm{a},\mathrm{b}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{3}. \\ $$$$\mathrm{Hence},\:\mathrm{3}\mid\left(\mathrm{ab}\right) \\ $$
Answered by vnm last updated on 22/Apr/25
let a=kp, b=kq, where p,q are coprime one of the numbers p,q can be represented as m^2 −n^2 and the other as 2mn if 3∣m or 3∣n then 3∣(2mn)⇒3∣(ab) if 3∤m and 3∤n then m=3r+s, n=3t+u where ∣s∣=∣u∣=1⇒3∣(m^2 −n^2 )⇒3∣(ab)
$${let}\:{a}={kp},\:{b}={kq},\:{where}\:{p},{q}\:{are}\:{coprime} \\ $$$${one}\:{of}\:{the}\:{numbers}\:{p},{q}\:{can}\:{be}\:{represented} \\ $$$${as}\:{m}^{\mathrm{2}} −{n}^{\mathrm{2}} \:{and}\:{the}\:{other}\:{as}\:\mathrm{2}{mn} \\ $$$${if}\:\mathrm{3}\mid{m}\:{or}\:\mathrm{3}\mid{n}\:{then}\:\mathrm{3}\mid\left(\mathrm{2}{mn}\right)\Rightarrow\mathrm{3}\mid\left({ab}\right) \\ $$$${if}\:\mathrm{3}\nmid{m}\:{and}\:\mathrm{3}\nmid{n} \\ $$$${then}\:{m}=\mathrm{3}{r}+{s},\:{n}=\mathrm{3}{t}+{u} \\ $$$${where}\:\mid{s}\mid=\mid{u}\mid=\mathrm{1}\Rightarrow\mathrm{3}\mid\left({m}^{\mathrm{2}} −{n}^{\mathrm{2}} \right)\Rightarrow\mathrm{3}\mid\left({ab}\right) \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *