I have a cryptographic question. First, you have the alphabet and space, comma and point and you translates it to numbers ($a=0,b=1,...,z=25,space=26,comma=27,point=28$). Then, you have to find a encryption that makes every answer unique, so every number from 0 to 28 gives a number and the outputs aren't the same .
There are two encryption strategies given and I have to prove that they are working (or not) and thus give a unique output (or not).
1) You raise the number of the letter you have to encrypt to the power of two and take the $\bmod 29 $ of that number, so the formula is: $$number^2 \equiv a \bmod29$$ and a is different for every number you put in. You have to prove that this isn't true because $$14^2 \bmod29 \equiv 15^2\bmod29$$ How can you prove it instead of just giving an example where it isn't true?
2) You raise the number to the power three and then take $\bmod29$, so the formula is: $$number^3 \equiv b \bmod29$$ and b is unique for every input. You have to prove this is true.
Are there numbers (f.e. primes) you put in the $\bmod$ for which this is always true: $$number^2 \equiv a \bmod x$$$$number^3 \equiv b \bmod x$$$$(0 \leqslant number < x)$$ and a and b are unique?
$(\pm1)^2 \equiv 1$ shows that $x^2 \bmod 29$ isn't unique. This works for every modulus greater than $2$.
$(n^3)^{19}\equiv n\bmod 29$ shows that $x^3 \bmod 29$ is unique: $$ x^3 \equiv y^3 \bmod 29 \implies x \equiv (x^3)^{19} \equiv (y^3)^{19} \equiv y \bmod 29 $$ This works for every modulus $m$ such that $\gcd(3,\phi(m))=1$.