$74x \equiv 1 \bmod 53$

Question

I've figured out that $x=-5$ for this to work but I don't know how to state the general solution? (all cases)

Anyone know? D:

Answer 1

Hint: If you have two solutions, $x_1$ and $x_2$, then $$ 74x_1\equiv74x_2\pmod{53} $$ Since $\gcd(53,74)=1$, what does $$ 53\mid74(x_1-x_2) $$ imply?

Answer 2

First, write the congruence as $$74x = 1 + 53y$$

where $y$ is any integer. Then, $$74x - 53y = 1$$

As you said before, $x = -5$ (and so $y = -7$, which works for congruence). Then, there are more solutions $x = -5 + 53k$ where $k$ is any integer.

Answer 3

Hint $\ $ If $\,a',a\,$ are solutions of $\,74x\equiv 1$ then $\,a'$ times $\,1\equiv 74a\,\Rightarrow\, a'\equiv\, \ldots\ $ Conversely $(\Leftarrow)$ multiplying $\, a'\equiv\, \ldots\,$ by $\,74\,$ shows $\,74a'\equiv 74(\dots)\,$ so if one of them is $\equiv 1\,$ then they both are.

Above we invoked a few times the fundamental Congruence Product Rule - see below. The converse works more generally: $\,a'\equiv a,\,\ f(a)\equiv 0\,\Rightarrow\, f(a')\equiv 0,\,$ for any polynomial $\,f(x)\,$ with integer coefficients, see the Polynomial Congruence Rule below.

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Congruence Sum Rule $\rm\qquad\quad A\equiv a,\quad B\equiv b\ \Rightarrow\ \color{#0a0}{A+B\,\equiv\, a+b}\ \ \ (mod\ m)$

Proof $\rm\ \ m\: |\: A\!-\!a,\ B\!-\!b\ \Rightarrow\ m\ |\ (A\!-\!a) + (B\!-\!b)\ =\ \color{#0a0}{A+B - (a+b)} $

Congruence Product Rule $\rm\quad\ A\equiv a,\ \ and \ \ B\equiv b\ \Rightarrow\ \color{#c00}{AB\equiv ab}\ \ \ (mod\ m)$

Proof $\rm\ \ m\: |\: A\!-\!a,\ B\!-\!b\ \Rightarrow\ m\ |\ (A\!-\!a)\ B + a\ (B\!-\!b)\ =\ \color{#c00}{AB - ab} $

Congruence Power Rule $\rm\quad\ A\equiv a\ \Rightarrow\ A^n\equiv a^n\ \ (mod\ m)$

Proof $\ $ It is true for $\rm\,n=1\,$ and $\rm\,A\equiv a,\ A^n\equiv a^n \Rightarrow\, A^{n+1}\equiv a^{n+1},\,$ by the Product Rule, so the result follows by induction on $\,n.$

Polynomial Congruence Rule $\ $ If $\,f(x)\,$ is polynomial with integer coefficients then $\ A\equiv a\ \Rightarrow\ f(A)\equiv f(a)\,\pmod m.$

Proof $\ $ By induction on $\, n = $ degree $f.\,$ Clear if $\, n = 0.\,$ Else $\,f(x) = f(0) + x\,g(x)\,$ for $\,g(x)\,$ a polynomial with integer coefficients of degree $< n.\,$ By induction $\,g(A)\equiv g(a)\,$ so $\, A g(A)\equiv a g(A)\,$ by the Product Rule. Hence $\,f(A) = f(0)+Ag(A)\equiv f(0)+ag(a) = f(a)\,$ by the Sum Rule.

Beware $ $ that such rules need not hold true for other operations, e.g. the exponential analog of above $\rm A^B\equiv a^b$ is not generally true (unless $\rm B = b,\,$ so it reduces to the Power Rule, so follows by inductively applying $\,\rm b\,$ times the Product Rule).

Answer 4

As $\displaystyle 74\equiv21, 74x\equiv1\pmod{53}\iff 21x\equiv1\iff x\equiv21^{-1}\equiv3^{-1}\cdot7^{-1}$

Now as $18\cdot3=54\equiv1\pmod{53}\iff3^{-1}\equiv18$

and $53\cdot2-7\cdot15=1\implies 7(-15)\equiv1\pmod{53}\iff7^{-1}\equiv-15$

$\displaystyle\implies x\equiv18(-15)\pmod{53}\equiv-5\equiv48$

So, we can write $x=-5+53a=48+53(a-1)=48+53b$ where $a,b$ are any integers