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Given RSA parameters $p=11$, $q=5$ and $e=13$ we have that $d=37$ using Euler's totient function.

Let $M\in Z_{55}$ and we have $(M^e)^d\mod{55}=M$.

Unexpectedly, having $d=7$ results in $33/55$ correctly decrypted messages! Even more interesting is that they follow a repeating pattern of two incorrectly messages followed by three correctly decrypted messages. No other private key is anywhere near this success rate. Except for $d=17$ (but I think that is due to Carmichael's totient function) and of course $d=37$.

Why does $d=7$ have such high success rate and why does it follow the repeating pattern?

Table showing only the unsuccessfully decrypted messages using $d=7$. You can see the pattern by looking at the $M$ sequence. \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline M & 2 & 3 & 7 & 8 & 12 & 13 & 17 & 18 & 22 & 23 & 27 & 28 & 32 & 33 & 37 & 38 & 42 & 43 & 47 & 48 & 52 & 53 \\ \hline (M^e)^d & 13 & 47 & 18 & 52 & 23 & 2 & 28 & 7 & 33 & 12 & 38 & 17 & 43 & 22 & 48 & 27 & 53 & 32 & 3 & 37 & 8 & 42 \\ \hline \end{array}

Background: This was found as $d=7$ was given as the answer, but I expect the author might have dropped the $3$ in $37$ and as $3/5$ entries are correct it might have slipped by.

Filip
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1 Answers1

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$\!\!\bmod 5\!:\ \ x\not\equiv 0\Rightarrow x^4\equiv 1\Rightarrow x^{90}\!-\!1 \equiv x^2\!-\!1\ $ with $2$ roots, so $x^{91}\!-\!x = x(x^{90}\!-\!1)$ has $3$ roots

$\!\!\bmod 11\!: x\not\equiv 0\Rightarrow x^{10}\!\equiv 1\Rightarrow x^{90}\!-\!1 \equiv 0 $ with $10$ roots, so $x^{91}\!-\!x = x(x^{90}\!-\!1)$ has $11$ roots

So by CRT the $3$ roots $\!\bmod 5$ and $11$ roots $\!\bmod 11$ combine to $\,\smash{\bbox[5px,border:1px solid #c00]{33}}$ roots of $\,x^{91}-x\pmod{\!55}$

Bill Dubuque
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