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Prove-that-3k-2-is-not-perfect-square-for-all-k-0-1-2-3-




Question Number 23269 by Rasheed.Sindhi last updated on 28/Oct/17
Prove that   3k+2 is not perfect square for  all k∈{0,1,2,3,...}.
$$\mathbb{P}\mathrm{rove}\:\mathrm{that} \\ $$$$\:\mathrm{3k}+\mathrm{2}\:\mathrm{is}\:\mathrm{not}\:\mathrm{perfect}\:\mathrm{square}\:\mathrm{for} \\ $$$$\mathrm{all}\:\mathrm{k}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},…\right\}. \\ $$
Answered by Tinkutara last updated on 28/Oct/17
Any number can be written as 3p,  3p+1 and 3p+2.  If it is 3p, it′s square=9p^2 =3k form  If it is 3p+1, it′s square=9p^2 +6p+1  =3k+1 form  If it is 3p+2, it′s square=9p^2 +12p+4  =3k+1 form  Hence a square cannot be in 3k+2  form.
$${Any}\:{number}\:{can}\:{be}\:{written}\:{as}\:\mathrm{3}{p}, \\ $$$$\mathrm{3}{p}+\mathrm{1}\:{and}\:\mathrm{3}{p}+\mathrm{2}. \\ $$$${If}\:{it}\:{is}\:\mathrm{3}{p},\:{it}'{s}\:{square}=\mathrm{9}{p}^{\mathrm{2}} =\mathrm{3}{k}\:{form} \\ $$$${If}\:{it}\:{is}\:\mathrm{3}{p}+\mathrm{1},\:{it}'{s}\:{square}=\mathrm{9}{p}^{\mathrm{2}} +\mathrm{6}{p}+\mathrm{1} \\ $$$$=\mathrm{3}{k}+\mathrm{1}\:{form} \\ $$$${If}\:{it}\:{is}\:\mathrm{3}{p}+\mathrm{2},\:{it}'{s}\:{square}=\mathrm{9}{p}^{\mathrm{2}} +\mathrm{12}{p}+\mathrm{4} \\ $$$$=\mathrm{3}{k}+\mathrm{1}\:{form} \\ $$$${Hence}\:{a}\:{square}\:{cannot}\:{be}\:{in}\:\mathrm{3}{k}+\mathrm{2} \\ $$$${form}. \\ $$
Commented by Rasheed.Sindhi last updated on 28/Oct/17
ThanX  a lot sir!
$$\mathbb{T}\mathrm{han}\mathcal{X}\:\:\mathrm{a}\:\mathrm{lot}\:\mathrm{sir}! \\ $$

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