Question: prove no integer $a$, $b$, $c$ exists if $12a - 13b = 11$
1. **State the problem:** We want to prove that there are no integers $a$ and $b$ such that the equation $$12a - 13b = 11$$ holds.
2. **Recall the key concept:** For a linear Diophantine equation of the form $$Ax + By = C$$ to have integer solutions, the greatest common divisor (gcd) of $A$ and $B$ must divide $C$.
3. **Calculate the gcd:**
$$\gcd(12, 13) = 1$$
4. **Check divisibility:** Since $1$ divides every integer, the gcd condition does not immediately rule out solutions.
5. **Use modular arithmetic to test:** Consider the equation modulo $13$:
$$12a - 13b \equiv 11 \pmod{13}$$
Since $13b \equiv 0 \pmod{13}$, this simplifies to:
$$12a \equiv 11 \pmod{13}$$
6. **Find the inverse of $12$ modulo $13$:**
Since $12 \equiv -1 \pmod{13}$, the inverse of $12$ modulo $13$ is $12$ itself because:
$$12 \times 12 = 144 \equiv 1 \pmod{13}$$
7. **Solve for $a$ modulo $13$:**
$$a \equiv 11 \times 12 = 132 \equiv 2 \pmod{13}$$
8. **Substitute back to find $b$:**
Rewrite the original equation:
$$12a - 13b = 11 \implies 13b = 12a - 11$$
For $b$ to be integer, $12a - 11$ must be divisible by $13$.
Using $a \equiv 2 \pmod{13}$, let $a = 13k + 2$:
$$12(13k + 2) - 11 = 156k + 24 - 11 = 156k + 13$$
Since $156k + 13$ is divisible by $13$, $b = \frac{156k + 13}{13} = 12k + 1$ is integer.
9. **Conclusion:** There exist infinitely many integer solutions $(a,b)$ of the form:
$$a = 13k + 2, \quad b = 12k + 1, \quad k \in \mathbb{Z}$$
Therefore, the statement "no integer $a$, $b$, $c$ exists if $12a - 13b = 11$" is false.
**Final answer:** Integer solutions do exist for the equation $12a - 13b = 11$.