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Question Number 94097 by Rio Michael last updated on 16/May/20
find all integers n  for which  13 ∣4(n^2 +1).
$$\mathrm{find}\:\mathrm{all}\:\mathrm{integers}\:{n}\:\:\mathrm{for}\:\mathrm{which}\:\:\mathrm{13}\:\mid\mathrm{4}\left({n}^{\mathrm{2}} +\mathrm{1}\right). \\ $$
Commented by Rasheed.Sindhi last updated on 17/May/20
This s equivalent to:                   13 ∣(n^2 +1)  ((n^2 +1)/(13))=k⇒13k−1=n^2   All k′s for which 13k−1 is  perfect square.  Some values of k:  k=2,5,25,34,74,89,...  n=5,8,18,21,31,34,44,...  General solutions:  n=13q+5,13q+8    ∀ q∈Z     .....  ...
$${This}\:{s}\:{equivalent}\:{to}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{13}\:\mid\left({n}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\frac{{n}^{\mathrm{2}} +\mathrm{1}}{\mathrm{13}}={k}\Rightarrow\mathrm{13}{k}−\mathrm{1}={n}^{\mathrm{2}} \\ $$$${All}\:{k}'{s}\:{for}\:{which}\:\mathrm{13}{k}−\mathrm{1}\:{is} \\ $$$${perfect}\:{square}. \\ $$$${Some}\:{values}\:{of}\:{k}: \\ $$$${k}=\mathrm{2},\mathrm{5},\mathrm{25},\mathrm{34},\mathrm{74},\mathrm{89},… \\ $$$${n}=\mathrm{5},\mathrm{8},\mathrm{18},\mathrm{21},\mathrm{31},\mathrm{34},\mathrm{44},… \\ $$$$\mathcal{G}{eneral}\:{solutions}: \\ $$$${n}=\mathrm{13}{q}+\mathrm{5},\mathrm{13}{q}+\mathrm{8}\:\:\:\:\forall\:{q}\in\mathbb{Z}\:\:\: \\ $$$$….. \\ $$$$… \\ $$
Answered by Rasheed.Sindhi last updated on 17/May/20
13 ∤ 4⇒13 ∣ (n^2 +1)  Let n=13k+i where k,i∈Z ∧ 0≤i≤12   n^2 +1=13^2 k^2 +2(13k)(i)+i^2 +1  13 ∣ (n^2 +1)⇒13∣{13^2 k^2 +2(13k)(i)+i^2 +1}        ⇒ 13 ∣ (i^2 +1)  In interval 0≤i≤12 only i=5,8  satisfy above.  Hence     n=13k+5,13k+8 are satisfying  values ∀ k∈Z
$$\mathrm{13}\:\nmid\:\mathrm{4}\Rightarrow\mathrm{13}\:\mid\:\left({n}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$${Let}\:{n}=\mathrm{13}{k}+{i}\:{where}\:{k},{i}\in\mathbb{Z}\:\wedge\:\mathrm{0}\leqslant{i}\leqslant\mathrm{12} \\ $$$$\:{n}^{\mathrm{2}} +\mathrm{1}=\mathrm{13}^{\mathrm{2}} {k}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{13}{k}\right)\left({i}\right)+{i}^{\mathrm{2}} +\mathrm{1} \\ $$$$\mathrm{13}\:\mid\:\left({n}^{\mathrm{2}} +\mathrm{1}\right)\Rightarrow\mathrm{13}\mid\left\{\mathrm{13}^{\mathrm{2}} {k}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{13}{k}\right)\left({i}\right)+{i}^{\mathrm{2}} +\mathrm{1}\right\} \\ $$$$\:\:\:\:\:\:\Rightarrow\:\mathrm{13}\:\mid\:\left({i}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$${In}\:{interval}\:\mathrm{0}\leqslant{i}\leqslant\mathrm{12}\:{only}\:{i}=\mathrm{5},\mathrm{8} \\ $$$${satisfy}\:{above}. \\ $$$${Hence}\: \\ $$$$\:\:{n}=\mathrm{13}{k}+\mathrm{5},\mathrm{13}{k}+\mathrm{8}\:{are}\:{satisfying} \\ $$$${values}\:\forall\:{k}\in\mathbb{Z} \\ $$
Answered by Rasheed.Sindhi last updated on 17/May/20
13 ∣4(n^2 +1) is aquivalent to:         13 ∣(n^2 +1)  ^• n:13k,13k+1,...,13k+4,13k+5,        13k+6,13k+7,13k+8,13k+9,        ...,13k+12  ^• For values of n written in black  We can easily prove that:                 13 ∤ (n^2 +1)      For example if n=13k+2      n^2 +1=13^2 k^2 +2.13k.2+5                 =13(13k^2 +4k)+5      ∴ 13 ∤ (n^2 +1)  ^• For values of n written in red  We can easily prove that:                 13 ∣ (n^2 +1)       For an example if n=13k+8       n^2 +1=13^2 k^2 +2.13k.8+65                  =13(13k^2 +16k+5)       ∴  13 ∣ (n^2 +1)  ^• Hence13 ∣ (n^2 +1) for only  13k+5 & 13k+8
$$\mathrm{13}\:\mid\mathrm{4}\left({n}^{\mathrm{2}} +\mathrm{1}\right)\:{is}\:{aquivalent}\:{to}: \\ $$$$\:\:\:\:\:\:\:\mathrm{13}\:\mid\left({n}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\:^{\bullet} {n}:\mathrm{13}{k},\mathrm{13}{k}+\mathrm{1},…,\mathrm{13}{k}+\mathrm{4},\mathrm{13}{k}+\mathrm{5}, \\ $$$$\:\:\:\:\:\:\mathrm{13}{k}+\mathrm{6},\mathrm{13}{k}+\mathrm{7},\mathrm{13}{k}+\mathrm{8},\mathrm{13}{k}+\mathrm{9}, \\ $$$$\:\:\:\:\:\:…,\mathrm{13}{k}+\mathrm{12} \\ $$$$\:^{\bullet} \mathcal{F}{or}\:{values}\:{of}\:{n}\:{written}\:{in}\:{black} \\ $$$${We}\:{can}\:{easily}\:{prove}\:{that}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{13}\:\nmid\:\left({n}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\:\:\:\:{For}\:{example}\:{if}\:{n}=\mathrm{13}{k}+\mathrm{2} \\ $$$$\:\:\:\:{n}^{\mathrm{2}} +\mathrm{1}=\mathrm{13}^{\mathrm{2}} {k}^{\mathrm{2}} +\mathrm{2}.\mathrm{13}{k}.\mathrm{2}+\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{13}\left(\mathrm{13}{k}^{\mathrm{2}} +\mathrm{4}{k}\right)+\mathrm{5} \\ $$$$\:\:\:\:\therefore\:\mathrm{13}\:\nmid\:\left({n}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\:^{\bullet} \mathcal{F}{or}\:{values}\:{of}\:{n}\:{written}\:{in}\:{red} \\ $$$${We}\:{can}\:{easily}\:{prove}\:{that}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{13}\:\mid\:\left({n}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\:\:\:\:\:{For}\:{an}\:{example}\:{if}\:{n}=\mathrm{13}{k}+\mathrm{8} \\ $$$$\:\:\:\:\:{n}^{\mathrm{2}} +\mathrm{1}=\mathrm{13}^{\mathrm{2}} {k}^{\mathrm{2}} +\mathrm{2}.\mathrm{13}{k}.\mathrm{8}+\mathrm{65} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{13}\left(\mathrm{13}{k}^{\mathrm{2}} +\mathrm{16}{k}+\mathrm{5}\right) \\ $$$$\:\:\:\:\:\therefore\:\:\mathrm{13}\:\mid\:\left({n}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\:^{\bullet} \mathcal{H}{ence}\mathrm{13}\:\mid\:\left({n}^{\mathrm{2}} +\mathrm{1}\right)\:{for}\:{only} \\ $$$$\mathrm{13}{k}+\mathrm{5}\:\&\:\mathrm{13}{k}+\mathrm{8} \\ $$
Commented by Rio Michael last updated on 17/May/20
thank you sir.
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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