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Exam question and proof

My main line of thinking goes as follows: (Modular Arithmetic)

(mod 2 ≡ 0) , ( mod 13 ≡ 0) => (mod 2*13 ≡ 0) which gives ( 0 * 0 = 0 ) => (mod 26 ≡ 0)

=> ( mod ( 2 * 13 )(2 * 13) ≡ 0 ) which gives ( (0 * 0)(0 * 0) = 0 ) => (mod 26^2 ≡ 0)

However the examiner states that it is correct for (mod 26 ≡ 0) but not for (mod 26^2 ≡ 0).

I simply do not see how my proof is not enough. (Exam is in swedish so hope its still understandable.)

Leo Långberg
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    I can't see how this relates to the image or that it proves anything. $a\mod b$ is meaningful to me, but what does $\mod b$ on its own mean? – Paul Aug 23 '23 at 10:10
  • You are trying to apply the fact that if a number is divisible by $26$ then it is also divisible by $26^2$. You can probably see that this is not true. It looks like you correctly proved that the number is divisible by $26$, but that's all. – Michal Adamaszek Aug 23 '23 at 10:32
  • By the way is it common in Sweden to write $X|Y$ meaning that $Y$ divides $X$ ? – Michal Adamaszek Aug 23 '23 at 10:33
  • $3^{3n+3}=27^{n+1}=(1+26)^{n+1}=1+26(n+1)+26^2k\equiv26n+27\bmod26^2$ – J. W. Tanner Aug 23 '23 at 13:36
  • Your argument only implies that it is divisible by $26$, not by $26^2$. See the linked dupe for correct proofs. – Bill Dubuque Aug 23 '23 at 15:33

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Your question is not entirely clear. As I understand it, you've shown that $$ 3^{3n+3}-26n-27\equiv0\pmod{2}\label{e1}\tag{1} $$ and $$ 3^{3n+3}-26n-27\equiv0\pmod{13}\label{e2}\tag{2} $$ for $\ n\ge0\ $, and are claiming that it follows from this that both $$ 3^{3n+3}-26n-27\equiv0\pmod{26}\label{e3}\tag{3} $$ and $$ 3^{3n+3}-26n-27\equiv0\pmod{26^2}\label{e4}\tag{4}\ . $$ If so, then your examiner is correct, congruence (\ref{e4}) does not follow just from congruences (\ref{e1}) and (\ref{e2}) alone. If $\ A=78\ $, for example, then $\ A\equiv0\pmod{2}\ $ and $\ A\equiv0\pmod{13}\ $, but $\ A\not\equiv0\pmod{26^2}\ $. To obtain congruence (\ref{e4}) using this sort of argument you would need to show that $$ 3^{3n+3}-26n-27\equiv0\pmod{2^2} $$ and $$ 3^{3n+3}-26n-27\equiv0\pmod{13^2}\ , $$ which is certainly doable, but neither of these conguences follow automatically from (\ref{e1}) or (\ref{e2}) just by themselves.

lonza leggiera
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