Problem on Analysis (Functions) Let f(x) be a function from reals to reals obeying the following: f(x) is continuous, f(0)=1, and f(m+n+1)=f(m)+f(n). Show that f(x)=1+x for all real numbers x.

I am a bit confused on how to start with the sum as I am new to analysis. Assuming f(x)=1+x. I can indeed show that it satisfies properties f(m+n+1)=f(m)+f(n), (where m and n are reals) by checking it for RHS and LHS alternatively but can't prove it using the condition itself
Suppose that $n = 0$. Thus, the formula becomes:

$$f(m + 1) = f(m) + f(0)$$
$$f(m+1) = f(m) + 1$$

Substituting $m = x$
$$f(x+1) = f(x) + 1$$

Differentiating both sides:
$$f'(x+1) = f'(x)$$

This clearly indicates that the function is a linear function (as for any other type the slope would be dependent on the value of $x$)

Thus:
$$f(x) = mx + c$$

Now we know that $f(0) = 1$. Substituting that value, we get $c = 1$.
Using: $$f(x+1) = f(x) + 1$$
And substituting $x = -1$ we get $f(-1) = 0$. Thus we get the value of $m = 1$.

Thus, the equation is:
$$f(x) = x + 1$$
Suppose that $n = 0$. Thus, the formula becomes:

$$f(m + 1) = f(m) + f(0)$$
$$f(m+1) = f(m) + 1$$

Substituting $m = x$
$$f(x+1) = f(x) + 1$$

Differentiating both sides:
$$f'(x+1) = f'(x)$$

This clearly indicates that the function is a linear function (as for any other type the slope would be dependent on the value of $x$)

Thus:
$$f(x) = mx + c$$

Now we know that $f(0) = 1$. Substituting that value, we get $c = 1$.
Using: $$f(x+1) = f(x) + 1$$
And substituting $x = -1$ we get $f(-1) = 0$. Thus we get the value of $m = 1$.

Thus, the equation is:
$$f(x) = x + 1$$