Question Number 2655 by Yozzi last updated on 24/Nov/15
$${Prove}\:{by}\:{contradiction}\:{that}\:{there} \\$$$${are}\:{no}\:{whole}\:{number}\:{solutions}\:\left({x},{y},{z}\right) \\$$$${to}\:{the}\:{equation}\:{z}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\$$$${where}\:{both}\:{x}\:{and}\:{y}\:{are}\:{odd}. \\$$
Answered by prakash jain last updated on 24/Nov/15
$$\mathrm{Let}\:\mathrm{us}\:\mathrm{assume}\:\mathrm{that}\:\mathrm{such}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{exists} \\$$$${j},{k},{l}\in\mathbb{N}\cup\left\{\mathrm{0}\right\} \\$$$${x}=\mathrm{2}{k}+\mathrm{1} \\$$$${y}=\mathrm{2}{j}+\mathrm{1} \\$$$$\mathrm{since}\:{x},{y}\:{are}\:{odd}\:{z}\:{must}\:{be}\:{even} \\$$$${z}=\mathrm{2}{l} \\$$$${z}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\$$$$\mathrm{4}{l}^{\mathrm{2}} =\mathrm{4}{k}^{\mathrm{2}} +\mathrm{4}{k}+\mathrm{1}+\mathrm{4}{j}^{\mathrm{2}} +\mathrm{4}{j}+\mathrm{1} \\$$$${l}^{\mathrm{2}} ={k}^{\mathrm{2}} +{k}+{j}^{\mathrm{2}} +{j}+\frac{\mathrm{1}}{\mathrm{2}} \\$$$${l}^{\mathrm{2}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{fraction}.\:\mathrm{Contradicts}\:\mathrm{that}\:{z}\:\mathrm{is}\:\mathrm{a}\:\mathrm{whole} \\$$$$\mathrm{number}\:\mathrm{or}\:{l}\:\mathrm{is}\:\mathrm{a}\:\mathrm{whole}\:\mathrm{number}. \\$$
Commented by Rasheed Soomro last updated on 24/Nov/15
$$\mathcal{I}{nspiring}! \\$$
Answered by Filup last updated on 24/Nov/15
$${z}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\$$$${z}\in\mathbb{Z}\:\mathrm{when}: \\$$$${odd}\:{x}=\mathrm{2}{t}+\mathrm{1}\:\:\:{t}\in\mathbb{Z} \\$$$${odd}\:{y}=\mathrm{2}{t}+\mathrm{3}\:\:\:{t}\in\mathbb{Z} \\$$$$\\$$$${z}^{\mathrm{2}} =\left(\mathrm{2}{t}+\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}{t}+\mathrm{3}\right)^{\mathrm{2}} \\$$$$=\left(\mathrm{4}{t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{1}+\left(\mathrm{4}{t}^{\mathrm{2}} +\mathrm{6}{t}+\mathrm{9}\right)\right) \\$$$$=\left(\mathrm{8}{t}^{\mathrm{2}} +\mathrm{8}{t}+\mathrm{10}\right) \\$$$$=\mathrm{2}\left(\mathrm{4}{t}\left({t}+\mathrm{1}\right)+\mathrm{5}\right) \\$$$$\\$$$${z}=\sqrt{\mathrm{2}\left(\mathrm{4}{t}\left({t}+\mathrm{1}\right)+\mathrm{5}\right)} \\$$$$=\sqrt{\mathrm{4}}\sqrt{\mathrm{2}{t}\left({t}+\mathrm{1}\right)+\frac{\mathrm{5}}{\mathrm{2}}} \\$$$${z}=\mathrm{2}\sqrt{\mathrm{2}{t}\left({t}+\mathrm{1}\right)+\frac{\mathrm{5}}{\mathrm{2}}}\:\:\:\mathrm{where}\:\:{t}\in\mathbb{Z} \\$$$$\mathrm{2}{t}\left({t}+\mathrm{1}\right)=\mathrm{whole}\:\mathrm{number} \\$$$$\\$$$$\\$$$${continue}? \\$$
Commented by Filup last updated on 24/Nov/15
$$\mathrm{I}\:\mathrm{just}\:\mathrm{realised}\:\mathrm{I}\:\mathrm{was}\:\mathrm{only}\:\mathrm{proving}\:\mathrm{for}\:{z}. \\$$$$\mathrm{I}\:\mathrm{mis}−\mathrm{read}\:\mathrm{the}\:\mathrm{question}.\:\mathrm{Didn}'\mathrm{t} \\$$$$\mathrm{realise}\:\mathrm{you}\:\mathrm{wanted}\:{x}\:\mathrm{and}\:{y},\:\mathrm{too}… \\$$
Commented by prakash jain last updated on 24/Nov/15
$$\mathrm{You}\:\mathrm{started}\:\mathrm{with}\:{x}=\mathrm{2}{t}+\mathrm{1}\:\mathrm{and}\:{y}=\mathrm{2}{t}+\mathrm{3} \\$$$$\mathrm{the}\:\mathrm{question}\:\mathrm{only}\:\mathrm{said}\:{x}\:\mathrm{and}\:{y}\:\mathrm{are}\:\mathrm{odd}\:\mathrm{and} \\$$$$\mathrm{did}\:\mathrm{not}\:\mathrm{state}\:\mathrm{that}\:{x}\:\mathrm{and}\:{y}\:\mathrm{are}\:\mathrm{consecutive} \\$$$$\mathrm{odd}\:\mathrm{numbers}. \\$$$$\mathrm{The}\:\mathrm{last}\:\mathrm{part}\:\mathrm{of}\:\mathrm{your}\:\mathrm{conclusion} \\$$$${z}=\mathrm{2}\sqrt{\mathrm{2}{t}\left({t}+\mathrm{1}\right)+\frac{\mathrm{5}}{\mathrm{2}}}\:\:\mathrm{is}\:\mathrm{valid}\:\mathrm{since}\:\mathrm{you}\:\mathrm{have} \\$$$$\mathrm{a}\:\mathrm{fraction}\:\mathrm{under}\:\mathrm{the}\:\mathrm{root}\:\mathrm{sign}\:\mathrm{so}\:\mathrm{you} \\$$$$\mathrm{cannot}\:\mathrm{get}\:\mathrm{a}\:\mathrm{whole}\:\mathrm{number}. \\$$$$\mathrm{Also}\:\mathrm{proving}\:\mathrm{for}\:{z}\:\mathrm{is}\:\mathrm{sufficient}\:\mathrm{as}\:\mathrm{the}\:\mathrm{question} \\$$$$\mathrm{only}\:\mathrm{requires}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{condition}\:\mathrm{on} \\$$$$\mathrm{all}\:\mathrm{3}\:\mathrm{variables}\:\mathrm{are}\:\mathrm{not}\:\mathrm{satisfied}\:\mathrm{simultaneously}. \\$$$$\mathrm{So}\:\mathrm{proving}\:\mathrm{by}\:\mathrm{assuming}\:\mathrm{2}\:\mathrm{variables}\:\mathrm{meet} \\$$$$\mathrm{condition}\:\mathrm{and}\:\mathrm{creating}\:\mathrm{a}\:\mathrm{contradiction}\:\mathrm{on} \\$$$$\mathrm{3rd}\:\mathrm{is}\:\mathrm{sufficient}. \\$$
Commented by Filup last updated on 24/Nov/15
$$\mathrm{2}{t}+\mathrm{1}\:\:\mathrm{and}\:\:\mathrm{2}{t}+\mathrm{3}\:\mathrm{are}\:\mathrm{odd},\:\mathrm{though}? \\$$
Commented by prakash jain last updated on 24/Nov/15
$$\mathrm{Yes}.\:\mathrm{But}\:\mathrm{consecutive}\:\mathrm{odd}\:\mathrm{number}.\:\mathrm{So} \\$$$$\mathrm{the}\:\mathrm{conditions}\:\mathrm{are}\:\mathrm{more}\:\mathrm{restrictive}\:\mathrm{than} \\$$$$\mathrm{required}\:\mathrm{by}\:\mathrm{question}. \\$$
Commented by Filup last updated on 25/Nov/15
$$\mathrm{I}\:\mathrm{understand}\:\mathrm{now} \\$$
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