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

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