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an-early-question-for-the-new-year-how-many-rectangular-triangles-with-sides-a-b-c-N-exist-with-one-side-2019-

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Question Number 46809 by MJS last updated on 31/Oct/18
an early question for the new year how many rectangular triangles with sides a, b, c ∈N^★ exist with one side =2019
$$\mathrm{an}\:\mathrm{early}\:\mathrm{question}\:\mathrm{for}\:\mathrm{the}\:\mathrm{new}\:\mathrm{year} \\ $$$$\mathrm{how}\:\mathrm{many}\:\mathrm{rectangular}\:\mathrm{triangles}\:\mathrm{with}\:\mathrm{sides} \\ $$$${a},\:{b},\:{c}\:\in\mathbb{N}^{\bigstar} \:\mathrm{exist}\:\mathrm{with}\:\mathrm{one}\:\mathrm{side}\:=\mathrm{2019} \\ $$
Answered by MrW3 last updated on 02/Nov/18
a^2 +b^2 =c^2 case 1: with c=2019 a^2 +b^2 =2019^2 =3^2 ×673^2 let a=3p, b=3q ⇒p^2 +q^2 =673^2 it has one and only one solution: p=385, q=552 ⇒a=3×385=1155 ⇒b=3×552=1656 case 2: with a=2019 2019^2 +b^2 =c^2 ⇒c^2 −b^2 =(c−b)(c+b)=2019^2 =3^2 ×673^2 =m×n=1×4076361=3×1358787=9×452929=673×6057 c−b=m c+b=n ⇒b=((n−m)/2) ⇒c=((n+m)/2) ⇒b=((4076361−1)/2)=2038180 ⇒c=((4076361+1)/2)=2038181 or ⇒b=((1358787−3)/2)=679392 ⇒c=((1358787+3)/2)=679395 or ⇒b=((452929−9)/2)=226460 ⇒c=((452929+9)/2)=226469 or ⇒b=((6057−673)/2)=2692 ⇒c=((6057+673)/2)=3365 so totally there exist 5 such triangles: a/b/c=1155/1656/2019 a/b/c=2019/2692/3365 a/b/c=2019/226460/226469 a/b/c=2019/679392/679395 a/b/c=2019/2038180/2038181
$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={c}^{\mathrm{2}} \\ $$$$\boldsymbol{{case}}\:\mathrm{1}:\:{with}\:{c}=\mathrm{2019} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{2019}^{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} ×\mathrm{673}^{\mathrm{2}} \\ $$$${let}\:{a}=\mathrm{3}{p},\:{b}=\mathrm{3}{q} \\ $$$$\Rightarrow{p}^{\mathrm{2}} +{q}^{\mathrm{2}} =\mathrm{673}^{\mathrm{2}} \\ $$$${it}\:{has}\:{one}\:{and}\:{only}\:{one}\:{solution}: \\ $$$${p}=\mathrm{385},\:{q}=\mathrm{552} \\ $$$$\Rightarrow{a}=\mathrm{3}×\mathrm{385}=\mathrm{1155} \\ $$$$\Rightarrow{b}=\mathrm{3}×\mathrm{552}=\mathrm{1656} \\ $$$$ \\ $$$$\boldsymbol{{case}}\:\mathrm{2}:\:{with}\:{a}=\mathrm{2019} \\ $$$$\mathrm{2019}^{\mathrm{2}} +{b}^{\mathrm{2}} ={c}^{\mathrm{2}} \\ $$$$\Rightarrow{c}^{\mathrm{2}} −{b}^{\mathrm{2}} =\left({c}−{b}\right)\left({c}+{b}\right)=\mathrm{2019}^{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} ×\mathrm{673}^{\mathrm{2}} ={m}×{n}=\mathrm{1}×\mathrm{4076361}=\mathrm{3}×\mathrm{1358787}=\mathrm{9}×\mathrm{452929}=\mathrm{673}×\mathrm{6057} \\ $$$${c}−{b}={m} \\ $$$${c}+{b}={n} \\ $$$$\Rightarrow{b}=\frac{{n}−{m}}{\mathrm{2}} \\ $$$$\Rightarrow{c}=\frac{{n}+{m}}{\mathrm{2}} \\ $$$$\Rightarrow{b}=\frac{\mathrm{4076361}−\mathrm{1}}{\mathrm{2}}=\mathrm{2038180} \\ $$$$\Rightarrow{c}=\frac{\mathrm{4076361}+\mathrm{1}}{\mathrm{2}}=\mathrm{2038181} \\ $$$${or} \\ $$$$\Rightarrow{b}=\frac{\mathrm{1358787}−\mathrm{3}}{\mathrm{2}}=\mathrm{679392} \\ $$$$\Rightarrow{c}=\frac{\mathrm{1358787}+\mathrm{3}}{\mathrm{2}}=\mathrm{679395} \\ $$$${or} \\ $$$$\Rightarrow{b}=\frac{\mathrm{452929}−\mathrm{9}}{\mathrm{2}}=\mathrm{226460} \\ $$$$\Rightarrow{c}=\frac{\mathrm{452929}+\mathrm{9}}{\mathrm{2}}=\mathrm{226469} \\ $$$${or} \\ $$$$\Rightarrow{b}=\frac{\mathrm{6057}−\mathrm{673}}{\mathrm{2}}=\mathrm{2692} \\ $$$$\Rightarrow{c}=\frac{\mathrm{6057}+\mathrm{673}}{\mathrm{2}}=\mathrm{3365} \\ $$$$ \\ $$$${so}\:\boldsymbol{{totally}}\:{there}\:{exist}\:\mathrm{5}\:{such}\:{triangles}: \\ $$$${a}/{b}/{c}=\mathrm{1155}/\mathrm{1656}/\mathrm{2019} \\ $$$${a}/{b}/{c}=\mathrm{2019}/\mathrm{2692}/\mathrm{3365} \\ $$$${a}/{b}/{c}=\mathrm{2019}/\mathrm{226460}/\mathrm{226469} \\ $$$${a}/{b}/{c}=\mathrm{2019}/\mathrm{679392}/\mathrm{679395} \\ $$$${a}/{b}/{c}=\mathrm{2019}/\mathrm{2038180}/\mathrm{2038181} \\ $$
Commented by MJS last updated on 02/Nov/18
you′re right, thank you!
$$\mathrm{you}'\mathrm{re}\:\mathrm{right},\:\mathrm{thank}\:\mathrm{you}! \\ $$

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