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

Prove that the sum $$\large S(n) = 1^{4n+2} + 2^{4n+2} + 3^{4n+2} + 4^{4n+2} + 5^{4n+2} + 6^{4n+2}$$ is divisible by 13 for all positive integer $$n$$.

Note by Aymen Hafeez
1 year, 1 month ago

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Consider modulo 13 $\begin{array}a 1^{4n+2}&\equiv &1(1^4)^n &\equiv&1(1)^n&\pmod{13}\\ 2^{4n+2}&\equiv &4(2^4)^n &\equiv &4(3)^n&\pmod{13}\\ 3^{3n+2}&\equiv &9(3^4)^n &\equiv &-4(3)^n&\pmod{13}\\ 4^{4n+2}&\equiv &16(4^4)^n&\equiv&3(9)^n &\pmod{13}\\ 5^{4n+2}&\equiv &25(5^4)^n&\equiv &-1(1)^n&\pmod{13}\\ 6^{4n+2}&\equiv &36(6^4)^n&\equiv &-3(9)^n&\pmod{13} \end{array}$ add them up to get zero · 1 year, 1 month ago

$$we\quad may\quad rewrite\quad S(n)\quad as\\ { 1 }^{ 2(2n+1) }\quad +\quad { 2 }^{ 2(2n+1) }\quad +\quad { 3 }^{ 2(2n+1) }\quad ....\quad upto\quad { 6 }^{ 2(2n+1) }\\ =\quad { 1 }^{ 2n+1 }\quad +\quad { 4 }^{ 2n+1 }\quad +\quad { 9 }^{ 2n+1 }\quad .....\quad upto\quad { 36 }^{ 2n+1 }\\ \\ now\quad using\quad the\quad property\quad that\quad a^ n\quad +\quad b^ n\quad is\quad always\quad divisible\quad by\quad a\quad +\quad b\quad when\quad n\quad is\quad odd\quad \\ and\quad also\quad the\quad fact\quad that\quad 2n+1\quad is\quad always\quad odd.\\ { 1 }^{ 2n+1 }\quad +\quad { 25 }^{ 2n+1 }\quad is\quad divisible\quad by\quad 26\quad and\quad hence\quad 13\quad -\quad (i)\\ { 4 }^{ 2n+1 }\quad +\quad { 9 }^{ 2n+1 }\quad is\quad divisible\quad by\quad 13-\quad (ii)\\ { 36 }^{ 2n+1 }\quad +\quad { 16 }^{ 2n+1 }\quad is\quad divisible\quad by\quad 52\quad and\quad hence\quad 13\quad -\quad (iii)\\ \quad adding\quad the\quad equations\quad up\quad we\quad get\quad \\ { 1 }^{ 2n+1 }\quad +\quad { 4 }^{ 2n+1 }....\quad upto\quad { 36 }^{ 2n+1 }\quad or\quad in\quad other\quad words\quad S(n)\quad will\quad always\quad be\quad divisible\quad by\quad 13\quad for\quad positive\quad integer\quad n\\ \\ hence\quad proved.\\$$ · 1 year, 1 month ago

A pretty easy and awesome solution! · 1 year ago