Another way to solve it is to use the Lifting the exponent lemma (LTE). So noting $p=17$ we need to solve $2^{a}+17^{b}=19^{a}=(2+17)^{a}>2^{a}+17^{a}\ \forall a>1$ $\therefore b\geq a+1$ or $a=1$ if the latter then $a=b=1$ otherwise: $17\mid19-2$ and $17\nmid19,17\nmid2$, $\therefore$ by LTE: $v_{17}(19^{a}-2^{a})=v_{17}(19-2)+v_{17}(a)$ $\Leftrightarrow b=1+v_{17}(a)$ $\therefore 1+v_{17}(a)\geq a+1$ $\Leftrightarrow v_{17}(a)\geq a$ which is untrue $\forall a \in Z>0$ So the only solution for $(a,b,p)$ is $(1,1,17)$

$p^b = 19^a-2^a$ If $a > 1$ then with Zsigmondy's Theorem there exists a prime p, such that $p \mid 19^a-2^a \wedge p \nmid 19-2 = 17$, so $a=1 \rightarrow p^b = 17 \rightarrow b=1 \wedge p=17$

$p^b=19^a-2^a\implies p=17$ since $p|19-2$ assume that $a=17^{c}\cdot d$ where $17\nmid d$ so with LTE $b=v_{17}(19^a-2^a)=v_{17}(19-2)+v_{17}(a)=1+c$ for $c\ge 1$ we have $17^{1+c}=19^a-2^a>19^{17^{c}}-2^{17^{c}}>(19-2)^{17^{c}}=17^{17^c}$ impossible so we get $c=0\implies b=1,a=1,p=17$

Easy using Zsigmondys theorem. We know that the theorem works for $a>1$ so we devide the problem in 2 cases. 1)$case1.$ $a=1$ from this be get that $p^b=17$ , which means that $p=17$. 2)$Case2$. $a>1$ from Zsigmondy's theorem we get that there exist a prime $p \mid 19^a-2^a \wedge p \nmid 19-2$ ,so $p\nmid 17$.

The original equation becomes $p^b=19^a-2^a=17(19^{a-1}+19^{a-2}\cdot 2+\dots +19\cdot 2^{a-2}+2^{a-1})$, which means that $17\mid p^b$: therefore $p=17$. Consequently, we have $2^a+17^b=19^a$. By analysing the equation $\textrm{mod}\,3$ we obtain $(-1)^a+(-1)^b\equiv 1^a\equiv 1$; in particular, this is true only if $a$ and $b$ are both odd. Then, looking again at the equation, this time $\textrm{mod}\,4$, we see that $2^a+1^b\equiv 2^a+1\equiv (-1)^a\equiv -1\Rightarrow 2^a\equiv -2\equiv 2\,(\textrm{mod}\,4)$ but of course this happens only if $a=1$ because if $a\geq 2$, then $2^a\equiv 0\,(\textrm{mod}\,4)$. Hence $17^b=19-2=17\Rightarrow b=1$ and so the only acceptable triple is $(1,1,17)$.

$19^a-2^a=p^b.$ Note that $17\mid p^b$ so $p=17$. We have $19^a-2^a=17^b$. Claim: $f(x)=19^x-2^x$ is an increasing function for $x\in\mathbb{N}$. Proof. Let $x>y>0$ be any positive integers. \begin{align*} f(x)>f(y) &\iff 19^x-2^x>19^y-2^y \\ &\iff 19^x-19^y>2^x-2^y \\ &\iff 19^y(19^{x-y}-1)>2^y(2^{x-y}-1) \\ \end{align*}which is true because $19^y>2^y>0$ and $19^{x-y}-1>2^{x-y}-1>0$. Case 1: $b=1$ $19^a-2^a=17$. Note that $19^1-2^1=17$. Since $f$ is an increasing function, there is only one solution such that $f(a)=17$. Case 2: $b>1$ \begin{align*} b-1 &= v_{17}(17^b)-1 \\ &= v_{17}(19^a-2^a)-1 \\ &= v_{17}(19-2)+v_{17}(a)-1 \\ &= 1 +v_{17}(a)-1 \\ &= v_{17}(a) \end{align*}We have $17^{b-1}\mid a\implies a\geq 17^{b-1}>0$. Thus, $17^b=19^a-2^a\geq 19^{17^{b-1}}-2^{17^{b-1}}$ because $f$ is an increasing function. \begin{align*} 17^b &=19^{17^{b-1}}-2^{17^{b-1}} \\ &= (19-2)\left(\sum_{i=0}^{17^{b-1}-1}19^{17^{b-1}-1-i}\cdot 2^i\right) \\ &\geq 17\left(\sum_{i=0}^{17^{b-1}-1}2^{17^{b-1}-1-i}\cdot 2^i\right) \\ &=17\left(17^{b-1}\cdot 2^{17^{b-1}-1}\right) \\ &> 17^b & \text{ which is absurd.} \end{align*}The only triple is $(a,b,p)=(1,1,17)$.

I hate induction... Solution: We will show $(a,p,b) = (1,17,1)$ is the only working solution. It clearly works since $2^1+17^1=19^1$. We will now show it is the only solution. See that \begin{align} 2^a+p^b=19^a \iff p^b = 19^a - 2^a \label{1} \end{align}From here, it is clear that $17 \mid 19^a-2^a$. This even establishes $p = 17$. From Lifting the Exponent Lemma we have \begin{align} b = \nu_{17}(19^a-2^a) = \nu_{17}(19-2)+\nu_{17}(a) = 1 + \nu_{17}(a) \tag{2} \end{align}Observe that $b=1$ gives us the claimed solution. We now show there are no possible values of $b>1$. Claim: $f(x) = 19^x-2^x$ is an increasing function over $\mathbb{R}^+$. Proof: Clearly $f(1)>0$. It just suffices $f'(x)>0$ for any $x \in \mathbb{R}^+$. This would also show that $f>0$ in $\mathbb{R}^+$. We now just compute the derivative and set it greater than 0. \begin{align*} \frac{df}{dx} = \ln(19)\cdot 19^a - \ln(2)\cdot 2^a>0 \iff (9.5)^a> \frac{\ln(2)}{\ln(19)} \end{align*}which is always true for $a \in \mathbb{R}^+$ because $\frac{\ln(2)}{\ln(19)}<1$ and any exponential function with base greater than 1 is always greater than 1 for positive real exponent. $\square$ Substitute from $(2)$ to $(1)$ to get \[17^{1+\nu_{17}(a)} = 19^a - 2^a\]The following lemma will finish the problem on the spot: Lemma: $19^a-2^a > 17^{1+\nu_{17}(a)}$ for $\nu_{17}(a)>1$ and $a \in \mathbb{Z}^+$. Proof: We will show this by induction on $\nu_{17}(a)$. We first show that the base case of $\nu_{17}(a) = 2$ holds true. The condition implies $17^2 \le a$. Since the function in the RHS is increasing, we only need to check $a = 289$ which is trivially true. Also, rewrite $a = 17^t\cdot m$ for $\gcd(17,m) = 1$ and $t = \nu_{17}(a)$. For the induction step assume the result to be true for for some $t \in \mathbb{Z}^+$. And we need to show this holds true for $t+1$ as well. We have to show that \[19^{17^{t+1}\cdot m} - 2^{17^{t+1}\cdot m} > 17^{t+2}\]From the induction hypothesis, we know \[19^{17^{t}\cdot m} - 2^{17^{t}\cdot m} > 17^{t+1}\]Multiplying both sides by $19-2 = 17$ gives \[19^{17^{t+1}\cdot m} - 2^{17^{t+1}\cdot m} >19^{17^t\cdot m +1} - 2\cdot 19^{17^t\cdot m} - 17\cdot 2^{17^t\cdot m} - 2^{t\cdot m +1} > 17^{t+2}\]which is true because $f$ is increasing and $t>1$ and the induction is complete. $\square$ The lemma is sufficient to conclude the problem due to size reasons. $\blacksquare$

Notice that $2^a \equiv 19^a \mod{17}$ So $0\equiv p \mod{17}, p=17$ Now suppose that $a$ is at least $2$ $17^b=19^a-2^a$ We know that $gcd(19,2)=1$, using Zsigmondy's theorem $\exists q:$ prime number $/ q \mid 19^a-2^a , q \nmid 19^c-2^c, \forall c \in \{ 1,2,\ldots n\}$ Then there exists $q$ prime number, $q\neq 17$ such that $q \mid 19^a-2^a$, but $q \mid 17^b=19^a-2^a$ (absurd!) So $a=1$, $2+17^b=19$ then $b=1$ Therefore $(a,b,p)=(1,1,17)$

WakeUp wrote: Find all triples of positive integers $(a,b,p)$ with $a,b$ positive integers and $p$ a prime number such that $2^a+p^b=19^a$ $\Rightarrow p^b=19^a-2^a$ $19-2=17$ $gcd(19,2)=1$ $\Rightarrow$By Zsigmondy's Theorem: $\exists$ prime $q\neq 17/ q|19^a-2^a, \forall a>1$ besides $19-2|19^a-2^a$ $\Rightarrow$ If $a>1, 19^a-2^a$ has two prime factors $(\Rightarrow\Leftarrow)$ $\Rightarrow a=1 \Rightarrow p=17,b=1$ $\Rightarrow (a,b,p)=(1,1,17)$ is the only solution$_\blacksquare$

Saw this as a quick exercise on DNW-expnt so here we go If a>1, 17\mid19^a-2^a=p^b\implies p=17; by Zsigmondy first form 17\mid19^a-2^a but not 19^k-2^k for k<a, which is a contradiction. Then a=1,b=1,p=17. This nowadays is like an AIME p5 for lengthwise but probably harder if you dont know zsigmondy

Notice that we can rewrite the equation as $p^b=19^a-2^a$ Moreover $\nu_{17}(p^b)=\nu_{17}(19^a-2^a)\overset{\text{LTE}}{=}\nu_{17}(17)+\nu_{17}(a)=\nu_{17}(a)+1\ge1$ thus since $p$ is prime and $\nu_{17}(p^b)\ge1$, $p$ must be equal to $17$ Substituting this result into our original equation we obtain $2^a+17^b=19^a$ Case 1: $17\nmid a$ $\nu_{17}(17^b)=1\Longrightarrow b=1$ plugging this in yields $17=19^a-2^a$ which forces $a=1$ since $19^a-2^a>17, \forall a>1$ which clearly satisfies the original equation Thus $(a,b,p)=(1,1,17)$ is one solution to the equation. Case 2: $17\mid a\Longleftrightarrow a=17k$ The equation transforms into $2^{17k}+17^b=19^a$ Inspecting $\pmod 4$ yields $2^{17k}+17b\equiv1\pmod 4$ furthermore $19^{17k}\equiv3^k\equiv(-1)^k\pmod 4$ which forces $k$ to be even. Now let $k=2n$ The equation becomes $2^{17\cdot 2n}+17^b=19^{17\cdot 2n}$ Furthermore looking at $\pmod 6$ we obtain $2^{17\cdot 2n}+17^b\equiv131072^{2n}+(-1)^b\equiv 2^{2n}+(-1)^b\equiv3\text{ or }5\pmod 6$ However notice that $19^{17\cdot 2n}\equiv 1\pmod 6$ which is clearly a contradiction Therefore there exist no solutions when $17\mid a$ In conclusion $\boxed{(a,b,p)=(1,1,17)\text{ is the only solution for }a,b\in\mathbb{Z}^+\text{ and }p\in\mathbb{P}}$ $\blacksquare$.

The only solution is $a = b = 1, p = 17$, which clearly works. Rewriting the equation as $p^b = 19^a - 2^a$. By LTE, $19^a - 2^a$ is divisible by $17$, so $p = 17$. By Zsigmondy, there is a prime dividing $19^b - 2^b$ that doesn't divide $19^a - 2^a$ for $b > a > 1$. So, $p \neq 17$, contradiction. Hence, the only solution is $(a, b, p) = (1, 1, 17)$.

The only solution is $\boxed{(a, b, p) = (1, 1, 17)}$. To see this note that we can rearrange to find, $$p^b = 19^a - 2^a = 17(\dots)$$and hence $p = 17$. Now by Zsigmondy's for $a > 1$ there exists a prime $p$ dividing $19^a - 2^a$, not dividing $19^k - 2^k$ for $k < a$. Hence there exists some $p$ dividing $19^a - 2^a$ not equal to $17$, contradiction.

The answer is $(a,b,p) = (1,1,17)$. Firstly note that $p^b = 19^a - 2^a \implies 17 \mid p^b \implies p = 17$. Now note that $\nu_{17}(17^b) = \nu_{17}(19^a - 2^a) = \nu_{17}(19 - 2) + \nu_{17}(a)\implies b = \nu_{17}(a) + 1$. This gives us that $b = \nu_{17}(a) + 1 \le \left\lfloor \log_{17} a \right\rfloor + 1 \le \log_{17} 17a$. Then we get $17^b \le 17^{\log_{17} 17a} = 17a$. This gives us that $19^a - 2^a \le 17a$. It is obvious that the LHS grows super fast and the RHS isn't that fast. The only condition where the inequality holds is when $a=1$. This gives us our only solution $(a,b,p) = (1,1,17)$ and we are done.

Note that $17=19-2\mid 19^a-2^a=p^b$, so $p=17$. If $a>1$ then $19^a-2^a$ has a prime factor other than $17$, by Zsigmondy. Thus $a=1$ and so $b=1$.

Zsigmondy and BooM !!

If $a>1$, then by Zsigmondy's Theorem, $\exists p\in \mathbb{P}$ s.t. $p\mid 19^a-2^a\wedge p\nmid 19-2=17$, therefore, $a=1$, and the only solution is $\boxed{(a,b,p)=(2,1,17)}$