The image contains a math exercise in French. Here's the translation and breakdown of the exercise:

Exercise 26:
Let $n \in \mathbb{N}$ (set of natural numbers). We define: $a = 2n + 1 \quad \text{and} \quad b = 9n + 4$

1. Determine if: $9a - 2b = 1$

2. Deduce that: $\gcd(a, b) = 1$

Solution Outline:

1. For Part 1:

   We have the given values for $a$ and $b$: $a = 2n + 1, \quad b = 9n + 4$

   Substitute $a$ and $b$ into the equation $9a - 2b$: $9a - 2b = 9(2n + 1) - 2(9n + 4)$ Simplify both terms: $9(2n + 1) = 18n + 9, \quad 2(9n + 4) = 18n + 8$ Therefore: $9a - 2b = (18n + 9) - (18n + 8) = 1$ Hence, the equation holds true.

2. For Part 2:

   The task is to deduce that $\gcd(a, b) = 1$, which means $a$ and $b$ are relatively prime.

   The equation $9a - 2b = 1$ is a form of the Bézout identity. Since the linear combination of $a$ and $b$ gives 1, it implies that the greatest common divisor (gcd) of $a$ and $b$ is 1. Hence, $\gcd(a, b) = 1$.

Would you like more details or have any questions on this? Here are 5 related questions for further exploration:

1. Can you generalize the solution to any linear combination $xa - yb = 1$?
2. How can Bézout’s identity help in proving the gcd of two numbers?
3. How does the gcd help in solving Diophantine equations like $9a - 2b = 1$?
4. What other methods can we use to compute $\gcd(a, b)$ apart from Bézout’s identity?
5. Can this approach be extended to numbers involving different coefficients, like $5a - 3b = 1$?

Tip:

In number theory, if a linear combination of two integers equals 1, the integers are coprime, i.e., their gcd is 1. This is a direct consequence of Bézout's theorem.