Question and Answers Forum

All Questions      Topic List

Number Theory Questions

Previous in All Question      Next in All Question      

Previous in Number Theory      Next in Number Theory      

Question Number 23105 by Rasheed.Sindhi last updated on 26/Oct/17

(∣m^2 −n^2 ∣,2mn,m^2 +n^2 ) is pythagorean triplet for all m,n∈N. This can be proved easily.Is the vice versa of the statement is also true? I-E If for a,b,c∈N ,a^2 +b^2 =c^2 then there exist m,n∈N such that m^2 +n^2 =c and {a,b}={∣m^2 −n^2 ∣,2mn}

$$\left(\mid\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} \mid,\mathrm{2mn},\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} \right)\:\mathrm{is}\:\mathrm{pythagorean} \\ $$$$\mathrm{triplet}\:\mathrm{for}\:\mathrm{all}\:\mathrm{m},\mathrm{n}\in\mathbb{N}.\:\mathrm{This}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{proved}\:\mathrm{easily}.\mathrm{Is}\:\mathrm{the}\:\mathrm{vice}\:\mathrm{versa}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{statement}\:\mathrm{is}\:\mathrm{also}\:\mathrm{true}? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{I}-\mathrm{E} \\ $$$$\mathrm{If}\:\:\mathrm{for}\:\mathrm{a},\mathrm{b},\mathrm{c}\in\mathbb{N}\:,\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} =\mathrm{c}^{\mathrm{2}} \:\mathrm{then}\:\mathrm{there}\: \\ $$$$\mathrm{exist}\:\mathrm{m},\mathrm{n}\in\mathbb{N}\:\mathrm{such}\:\mathrm{that}\:\mathrm{m}^{\mathrm{2}} +\mathrm{n}^{\mathrm{2}} =\mathrm{c}\:\mathrm{and} \\ $$$$\left\{\mathrm{a},\mathrm{b}\right\}=\left\{\mid\mathrm{m}^{\mathrm{2}} −\mathrm{n}^{\mathrm{2}} \mid,\mathrm{2mn}\right\} \\ $$

Commented by prakash jain last updated on 27/Oct/17

(m^2 −n^2 )^2 +4m^2 n^2 =(m^2 +n^2 )^2 a=m^2 −n^2 b=4m^2 n^2 a=m^2 −n^2 c=m^2 +n^2 m=(√((c+a)/2)), n=(√((c−a)/2)) case (i) Let us say a and c are odd and b is even. c=2k+1 a=2l+1 c^2 −a^2 =4(k^2 −l^2 )+4(k−l) =4(k−l)(k+l−1) c+a=2(k+l+1) c−a=2(k−l) b=2(√((k−l)(k+l+1))) a=(m^2 −n^2 ),b=4m^2 n^2 if to prove: (k−l) and (k+l+1) are both whole squares. it is easy to see if (k−l) is whole square then (k+l+1) has to be a whole square since b is an integer. assume k−l=h^2 p where p is not whole square then (k+l+1)=j^2 p k=((j^2 p+h^2 p−1)/2), l=((j^2 p−h^2 p−1)/2) ...(A) We can easily see below if p exists in (A) then p is a common factor of a,b,c k=((j^2 p+h^2 p−1)/2), l=((j^2 p−h^2 p−1)/2) ...(A) a=2l+1=(j^2 −h^2 )p c=2k+1=(j^2 +h^2 )p So p exists as in (A) then p is a common factor of (a,b,c) Summary Every pythagorean triple (a,b,c) can be written in the form (m^2 −n^2 ,2mn,m^2 −n^2 ) if (a,b,c) have no common factor where m=(√((c−a)/2)),n=(√((c+a)/2))

$$\left({m}^{\mathrm{2}} −{n}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}{m}^{\mathrm{2}} {n}^{\mathrm{2}} =\left({m}^{\mathrm{2}} +{n}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$${a}={m}^{\mathrm{2}} −{n}^{\mathrm{2}} \\ $$$${b}=\mathrm{4}{m}^{\mathrm{2}} {n}^{\mathrm{2}} \\ $$$${a}={m}^{\mathrm{2}} −{n}^{\mathrm{2}} \\ $$$${c}={m}^{\mathrm{2}} +{n}^{\mathrm{2}} \\ $$$${m}=\sqrt{\frac{{c}+{a}}{\mathrm{2}}},\:{n}=\sqrt{\frac{{c}−{a}}{\mathrm{2}}} \\ $$$${case}\:\left({i}\right) \\ $$$$\mathrm{Let}\:\mathrm{us}\:\mathrm{say}\:{a}\:{and}\:{c}\:{are}\:{odd}\:\mathrm{and}\:{b}\:\mathrm{is}\:\mathrm{even}. \\ $$$${c}=\mathrm{2}{k}+\mathrm{1} \\ $$$${a}=\mathrm{2}{l}+\mathrm{1} \\ $$$${c}^{\mathrm{2}} −{a}^{\mathrm{2}} =\mathrm{4}\left({k}^{\mathrm{2}} −{l}^{\mathrm{2}} \right)+\mathrm{4}\left({k}−{l}\right) \\ $$$$=\mathrm{4}\left({k}−{l}\right)\left({k}+{l}−\mathrm{1}\right) \\ $$$${c}+{a}=\mathrm{2}\left({k}+{l}+\mathrm{1}\right) \\ $$$${c}−{a}=\mathrm{2}\left({k}−{l}\right) \\ $$$${b}=\mathrm{2}\sqrt{\left({k}−{l}\right)\left({k}+{l}+\mathrm{1}\right)} \\ $$$${a}=\left({m}^{\mathrm{2}} −{n}^{\mathrm{2}} \right),{b}=\mathrm{4}{m}^{\mathrm{2}} {n}^{\mathrm{2}} \:\mathrm{if} \\ $$$$\mathrm{to}\:\mathrm{prove}:\:\left({k}−{l}\right)\:\mathrm{and}\:\left({k}+{l}+\mathrm{1}\right)\:\mathrm{are}\:\mathrm{both}\:\mathrm{whole} \\ $$$$\mathrm{squares}. \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{see}\:\mathrm{if}\:\left({k}−{l}\right)\:\mathrm{is}\:\mathrm{whole} \\ $$$$\mathrm{square}\:\mathrm{then}\:\left({k}+{l}+\mathrm{1}\right)\:\mathrm{has}\:\mathrm{to}\:\mathrm{be} \\ $$$$\mathrm{a}\:\mathrm{whole}\:\mathrm{square}\:\mathrm{since}\:{b}\:\mathrm{is}\:\mathrm{an}\:\mathrm{integer}. \\ $$$$\mathrm{assume}\:{k}−{l}={h}^{\mathrm{2}} {p}\:\mathrm{where}\:{p}\:\mathrm{is}\:\mathrm{not} \\ $$$$\mathrm{whole}\:\mathrm{square}\:\mathrm{then}\:\left({k}+{l}+\mathrm{1}\right)={j}^{\mathrm{2}} {p} \\ $$$${k}=\frac{{j}^{\mathrm{2}} {p}+{h}^{\mathrm{2}} {p}−\mathrm{1}}{\mathrm{2}},\:{l}=\frac{{j}^{\mathrm{2}} {p}−{h}^{\mathrm{2}} {p}−\mathrm{1}}{\mathrm{2}}\:\:...\left({A}\right) \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{easily}\:\mathrm{see}\:\mathrm{below}\:\mathrm{if}\:{p} \\ $$$${exists}\:{in}\:\left({A}\right)\:{then}\:{p}\:{is}\:{a}\:{common} \\ $$$${factor}\:\mathrm{of}\:{a},{b},{c} \\ $$$${k}=\frac{{j}^{\mathrm{2}} {p}+{h}^{\mathrm{2}} {p}−\mathrm{1}}{\mathrm{2}},\:{l}=\frac{{j}^{\mathrm{2}} {p}−{h}^{\mathrm{2}} {p}−\mathrm{1}}{\mathrm{2}}\:\:...\left({A}\right) \\ $$$${a}=\mathrm{2}{l}+\mathrm{1}=\left({j}^{\mathrm{2}} −{h}^{\mathrm{2}} \right){p} \\ $$$${c}=\mathrm{2}{k}+\mathrm{1}=\left({j}^{\mathrm{2}} +{h}^{\mathrm{2}} \right){p} \\ $$$$\mathrm{So}\:{p}\:\mathrm{exists}\:\mathrm{as}\:\mathrm{in}\:\left({A}\right)\:\mathrm{then}\:{p}\:\mathrm{is}\:\mathrm{a}\:\mathrm{common} \\ $$$$\mathrm{factor}\:\mathrm{of}\:\left({a},{b},{c}\right) \\ $$$$\boldsymbol{\mathrm{Summary}} \\ $$$$\mathrm{Every}\:\mathrm{pythagorean}\:\mathrm{triple}\:\left({a},{b},{c}\right) \\ $$$$\mathrm{can}\:\mathrm{be}\:\mathrm{written}\:\mathrm{in}\:\mathrm{the}\:\mathrm{form} \\ $$$$\left({m}^{\mathrm{2}} −{n}^{\mathrm{2}} ,\mathrm{2}{mn},{m}^{\mathrm{2}} −{n}^{\mathrm{2}} \right)\:\mathrm{if}\:\left({a},{b},{c}\right) \\ $$$$\mathrm{have}\:\mathrm{no}\:\mathrm{common}\:\mathrm{factor} \\ $$$$\mathrm{where}\:{m}=\sqrt{\frac{{c}−{a}}{\mathrm{2}}},{n}=\sqrt{\frac{{c}+{a}}{\mathrm{2}}} \\ $$

Commented by prakash jain last updated on 26/Oct/17

case (ii) c even and a and b odd a=2k+1 b=2i+1 c=2l (2k+1)^2 +(2i+1)^2 =4l^2 4k^2 +4k+1+4i^2 +4k+1=4l^2 k^2 +k+i^2 +i+(1/2)=l^2 not possible

$${case}\:\left({ii}\right) \\ $$$${c}\:{even}\:{and}\:{a}\:{and}\:{b}\:{odd} \\ $$$${a}=\mathrm{2}{k}+\mathrm{1} \\ $$$${b}=\mathrm{2}{i}+\mathrm{1} \\ $$$${c}=\mathrm{2}{l} \\ $$$$\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}{i}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4}{l}^{\mathrm{2}} \\ $$$$\mathrm{4}{k}^{\mathrm{2}} +\mathrm{4}{k}+\mathrm{1}+\mathrm{4}{i}^{\mathrm{2}} +\mathrm{4}{k}+\mathrm{1}=\mathrm{4}{l}^{\mathrm{2}} \\ $$$${k}^{\mathrm{2}} +{k}+{i}^{\mathrm{2}} +{i}+\frac{\mathrm{1}}{\mathrm{2}}={l}^{\mathrm{2}} \\ $$$${not}\:{possible} \\ $$

Commented by Rasheed.Sindhi last updated on 27/Oct/17

Sir thanks for a nice answer! BTW Sir, why do you seem very rarely in the forum-activities nowadays?

$$\boldsymbol{\mathrm{Sir}}\:\mathrm{thanks}\:\mathrm{for}\:\mathrm{a}\:\mathrm{nice}\:\mathrm{answer}! \\ $$$$\mathrm{BTW}\:\mathrm{Sir},\:\mathrm{why}\:\mathrm{do}\:\mathrm{you}\:\mathrm{seem}\:\mathrm{very}\:\mathrm{rarely} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{forum}-\mathrm{activities}\:\mathrm{nowadays}? \\ $$

Commented by prakash jain last updated on 27/Oct/17

Past few months i was very busy in my office. So did not get any time at all.

$$\mathrm{Past}\:\mathrm{few}\:\mathrm{months}\:\mathrm{i}\:\mathrm{was}\:\mathrm{very}\:\mathrm{busy} \\ $$$$\mathrm{in}\:\mathrm{my}\:\mathrm{office}.\:\mathrm{So}\:\mathrm{did}\:\mathrm{not}\:\mathrm{get}\:\mathrm{any} \\ $$$$\mathrm{time}\:\mathrm{at}\:\mathrm{all}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com