Prove S(n)S(n) is divisible by 13

Prove that the sum S(n)=14n+2+24n+2+34n+2+44n+2+54n+2+64n+2\large S(n) = 1^{4n+2} + 2^{4n+2} + 3^{4n+2} + 4^{4n+2} + 5^{4n+2} + 6^{4n+2} is divisible by 13 for all positive integer nn.

#NumberTheory

Note by Aymen Hafeez
5 years, 5 months ago

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Comments

Consider modulo 13 a14n+2≡1(14)n≡1(1)n(mod13)24n+2≡4(24)n≡4(3)n(mod13)33n+2≡9(34)n≡−4(3)n(mod13)44n+2≡16(44)n≡3(9)n(mod13)54n+2≡25(54)n≡−1(1)n(mod13)64n+2≡36(64)n≡−3(9)n(mod13)\begin{array}{c}a 1^{4n+2}&\equiv &1(1^4)^n &\equiv&1(1)^n&\pmod{13}\\ 2^{4n+2}&\equiv &4(2^4)^n &\equiv &4(3)^n&\pmod{13}\\ 3^{3n+2}&\equiv &9(3^4)^n &\equiv &-4(3)^n&\pmod{13}\\ 4^{4n+2}&\equiv &16(4^4)^n&\equiv&3(9)^n &\pmod{13}\\ 5^{4n+2}&\equiv &25(5^4)^n&\equiv &-1(1)^n&\pmod{13}\\ 6^{4n+2}&\equiv &36(6^4)^n&\equiv &-3(9)^n&\pmod{13} \end{array} add them up to get zero

Aareyan Manzoor - 5 years, 5 months ago

wemayrewriteS(n)as12(2n+1)+22(2n+1)+32(2n+1)....upto62(2n+1)=12n+1+42n+1+92n+1.....upto362n+1nowusingthepropertythatan+bnisalwaysdivisiblebya+bwhennisoddandalsothefactthat2n+1isalwaysodd.12n+1+252n+1isdivisibleby26andhence13−(i)42n+1+92n+1isdivisibleby13−(ii)362n+1+162n+1isdivisibleby52andhence13−(iii)addingtheequationsupweget12n+1+42n+1....upto362n+1orinotherwordsS(n)willalwaysbedivisibleby13forpositiveintegernhenceproved.we\quad may\quad rewrite\quad S(n)\quad as\\ { 1 }^{ 2(2n+1) }\quad +\quad { 2 }^{ 2(2n+1) }\quad +\quad { 3 }^{ 2(2n+1) }\quad ....\quad upto\quad { 6 }^{ 2(2n+1) }\\ =\quad { 1 }^{ 2n+1 }\quad +\quad { 4 }^{ 2n+1 }\quad +\quad { 9 }^{ 2n+1 }\quad .....\quad upto\quad { 36 }^{ 2n+1 }\\ \\ now\quad using\quad the\quad property\quad that\quad a^ n\quad +\quad b^ n\quad is\quad always\quad divisible\quad by\quad a\quad +\quad b\quad when\quad n\quad is\quad odd\quad \\ and\quad also\quad the\quad fact\quad that\quad 2n+1\quad is\quad always\quad odd.\\ { 1 }^{ 2n+1 }\quad +\quad { 25 }^{ 2n+1 }\quad is\quad divisible\quad by\quad 26\quad and\quad hence\quad 13\quad -\quad (i)\\ { 4 }^{ 2n+1 }\quad +\quad { 9 }^{ 2n+1 }\quad is\quad divisible\quad by\quad 13-\quad (ii)\\ { 36 }^{ 2n+1 }\quad +\quad { 16 }^{ 2n+1 }\quad is\quad divisible\quad by\quad 52\quad and\quad hence\quad 13\quad -\quad (iii)\\ \quad adding\quad the\quad equations\quad up\quad we\quad get\quad \\ { 1 }^{ 2n+1 }\quad +\quad { 4 }^{ 2n+1 }....\quad upto\quad { 36 }^{ 2n+1 }\quad or\quad in\quad other\quad words\quad S(n)\quad will\quad always\quad be\quad divisible\quad by\quad 13\quad for\quad positive\quad integer\quad n\\ \\ hence\quad proved.\\

Shreyash Rai - 5 years, 4 months ago

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A pretty easy and awesome solution!

Adarsh Kumar - 5 years, 4 months ago

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Thanks!

Shreyash Rai - 5 years, 4 months ago
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