35. Find the positive integer tuples $(a, b, x, y)$ that satisfy the following condition:
$$(a+b)^{x}=a^{y}+b^{y}$$
35. When $a=b=1$, it is easy to see that the solution is $(a, b, x, y)=(1,1,1, y)$, where $y$ is any positive integer.

When $y=1$, the solution is $(a, b, 1,1)$, where $a, b$ are any positive integers.
When $y=2$, it is known that $x=1$, at this time $a+b=a^{2}+b^{2}$, so $a(a-1)=b(1-b) \geqslant 0$, which gives $a=b=1$.

When $y=3$, it is known that $x2^{\frac{3}{2}}$, we can get $2^{y^{-1}}>2^{\frac{3}{2}(t+k)\left(y^{-x}\right)}$, i.e.,
$$y-1>\frac{3}{2}(t+k)(y-x)$$

Now compare the exponents of the prime factorization of $p$ on both sides of (19), we should have
$$\begin{aligned}
v_{p}\left(a_{1}^{y}+b_{1}^{\gamma}\right) & =k x-t(y-x)>k x-\frac{2}{3}\left((y-1)-\frac{3}{2} k(y-x)\right) \\
& =k y-\frac{2}{3}(y-1)
\end{aligned}$$

Here we use (21).
On the other hand, note that $\left(a_{1}, b_{1}\right)=1$, so $\left(a_{1}, a_{1}+b_{1}\right)=1$. If $y$ is even, then $a_{1}^{Y}+b_{1}^{Y} \equiv 2 a_{1}^{Y} \neq 0(\bmod p)$, i.e., $p \nmid\left(a_{1}^{Y}+b_{1}^{Y}\right)$, which contradicts (22), so $y$ is odd. At this time, set $b_{1}=p^{k} \cdot u-a_{1}, u \in \mathbf{N}^{*}, p \nmid u$, then using the binomial theorem we know
$$\begin{aligned}
a_{1}^{y}+b_{1}^{y} & =a_{1}^{y}+\left(p^{k} \cdot u-a_{1}\right)^{y} \\
& =\sum_{n=2}^{y}(-1)^{n-1} C_{y}^{n} p^{n k} u^{n} a_{1}^{y-n}+y \cdot p^{k} \cdot u \cdot a_{1}^{y-1} .
\end{aligned}$$

Set $p^{s} \| y, s \in \mathbf{N}$. It is known that when $n \geqslant 2$, we have
$$\begin{aligned}
v_{p}\left(\mathrm{C}_{y}^{n} \cdot p^{n k} \cdot u^{n} \cdot a_{1}^{y^{-n}}\right) & =v_{p}\left(\frac{y}{n} \mathrm{C}_{y}^{n-1} \cdot p^{*}\right) \\
& \geqslant s+n k-v_{p}(n) \geqslant s+k+1
\end{aligned}$$

Here we use $p \geqslant 3, n \geqslant 3$ when, $p^{v} p^{(n)} \leqslant n \leqslant p^{n-2}$, so $v_{p}(n) \leqslant n-2$ (note that when $n=2$, $v_{p}(n)=0$, this inequality also holds).

And $v_{p}\left(y \cdot p^{k} \cdot u \cdot a_{1}^{Y-1}\right)=s+k$. Therefore, $v_{p}\left(a_{1}^{Y}+b_{1}^{y}\right)=s+k$. Thus, combining (22) we have
$$s+k>k y-\frac{2}{3}(y-1)$$

Thus, $s>k(y-1)-\frac{2}{3}(y-1) \geqslant \frac{1}{3}(y-1)$, so $s \geqslant\left[\frac{y-1}{3}\right]+1=$ $\left[\frac{y+2}{3}\right]$. Therefore, $p^{\left[\frac{2+2}{3}\right]} \leqslant p^{s} \leqslant y$, but this inequality does not hold when $p \geqslant 3, y \geqslant 4$.

The above discussion shows that $a_{1}+b_{1}$ can only have one prime factor 2. Set $a_{1}+b_{1}=2^{k}$, combining $\left(a_{1}, b_{1}\right)=1$, we know that $a_{1}, b_{1}$ are both odd. Using (19) we can set $d=2^{2}$, then
$$a_{1}^{y}+b_{1}^{y}=2^{k-x(y-x)} .$$

If $k=1$, then $a_{1}=b_{1}=1$, we get $a=b=2^{t}$, and require $x-t(y-x)=$ 1, solving this linear Diophantine equation in $x, y$ (for fixed $t$), we get the solution $(a, b, x, y)=$ $\left(2^{t}, 2^{t}, t s+1,(t+1) s+1\right), t, s \in \mathbf{N}$.

If $k \geqslant 2$, similarly analyzing the modulus 4 of both sides of (23), we know that $y$ is odd, at this time we have
$$a_{1}^{y}+b_{1}^{y}=a_{1}^{y}+\left(2^{k}-a_{1}\right)^{y}=\sum_{n=2}^{y}(-1)^{n-1} C_{y}^{n} 2^{2^{k}} a_{1}^{y-n}+y \cdot 2^{k} \cdot a_{1}^{y-1} .$$

Similarly analyzing the power of 2 on both sides of the above equation, we get $2^{k} \|\left(a_{1}^{y}+b_{1}^{*}\right)$, by (23) we have $a_{1}^{\gamma}+b_{1}^{y}=2^{k}$, thus $y=1$, which is a contradiction.

In summary, all positive integer solutions to the equation are $(a, b, x, y)=(1,1,1, y),(a, b$, $1,1),(2,1,2,3),(1,2,2,3),(2,2,2,3)$ or $\left(2^{t}, 2^{t}, t s+1,(t+1) s+1\right)$, where $a, b, y \in \mathbf{N}^{*}, t, s \in \mathbf{N}$.