Prove that at least two are the same $p = b^c + a , q = a^b + c , r = c^a + b$

Question

Given a, b, c ∈ N
p = b^c + a ,
q = a^b + c ,
r = c^a + b
we know that p q r are primes. Prove that at least two of the p ,q ,r are the same.

Edit: i have tried with contradiction method.I assumed all three p q r were not the same got 2 cases where i have 2 even numbers and one odd which is contadiction but that is dead end because after i can only get one even and two odd numbers which doesn't tell me anything

Answer 1

Assume $a,b,c$ are positive integers and $p,q,r$ are prime numbers such that \begin{align*} p&=c^a+b\\[4pt] q&=a^b+c\\[4pt] r&=b^c+a\\[4pt] \end{align*}

Claim:$\;$At least two of $p,q,r$ must be equal.

Suppose instead that $p,q,r$ are all distinct

If two of $p,q,r$ are even, those two primes must both be equal to $2$, contradiction.

Hence at most of one of $p.q.r$ is even.

Consider four cases . . .

Case $(1)$:$\;p$ is even and $q,r$ are both odd. \begin{align*} & p\;\text{is even} \\[4pt] \implies\;& p=2 \\[4pt] \implies\;& b=a=1 \\[4pt] \implies\;& q=1+c\;\text{and}\;r=c+1 \\[4pt] \implies\;& q=r \\[4pt] \end{align*} contradiction.

Case $(2)$:$\;q$ is even and $p,r$ are both odd. \begin{align*} & q\;\text{is even} \\[4pt] \implies\;& q=2 \\[4pt] \implies\;& a=c=1 \\[4pt] \implies\;& p=b+1\;\text{and}\;r=1+b \\[4pt] \implies\;& p=r \\[4pt] \end{align*} contradiction.

Case $(3)$:$\;r$ is even and $p,q$ are both odd. \begin{align*} & r\;\text{is even} \\[4pt] \implies\;& r=2 \\[4pt] \implies\;& c=b=1 \\[4pt] \implies\;& p=1+a\;\text{and}\;q=a+1 \\[4pt] \implies\;& p=q \\[4pt] \end{align*} contradiction.

Case $(4)$:$\;p,q,r$ are all odd.

First suppose $a$ is even.

• Since $a$ is even and $p$ is odd, $b$ must be odd.
• Then since $b$ is odd and $r$ is odd, $c$ must be even.
• But then since $c$ is even and $q$ is odd, $a$ must be odd, contradiction.

Next suppose $a$ is odd.

• Since $a$ is odd and $p$ is odd, $b$ must be even.
• Then since $b$ is even and $r$ is odd, $c$ must be odd.
• But then since $c$ is odd and $q$ is odd, $a$ must be even, contradiction.

Thus each of the four cases yields a contradiction.

It follows that $p,q,r$ cannot all be distinct, hence at least two of $p,q,r$ must be equal.

Answer 2

Since $p$ is odd by definition (assuming it isn't equal to $2$), then either $a$ is odd and $b$ and $c$ are even, or vice-versa.

Suppose $a$ is odd and $b$ and $c$ are even. Then $q$ is odd, but $r$ is even. Since $r$ is prime, it must be equal to $2$. But $r=c^a+b$, and the only solutions to $c^a+b=2$ such that $a$ is odd and $b$ and $c$ are even is $a=a, b=2, c=0$. We now check to see whether these values result in $p$ and $q$ being prime:

$p=b^c+a=2^0+a=1+a$

$q=a^b+c=a^2+0=a^2$

As we can see, whatever the value of $a$, $q$ cannot be prime. It must therefore be the case that $a$ is even and that $b$ and $c$ are odd. So again we look at $r=c^a+b$. An odd number raised to an even number is odd and the sum of two odd numbers is even, meaning that $r$ must once again be equal to $2$. This time, the only solution is $a=0, b=1, c=c$. I have left $c$ as an indefinite value as any odd value of $c$ will satisfy this equation. We therefore turn our attention to $p$ and $q$:

$p=b^c+a = 1^c+0=1$

$q=a^b+c=0^1+c=c$

Okay, that's no good either. Since neither case — two odd and one even, or two even and one odd — results in all three of $p,q,r$ being prime, it must be that all three of $a,b,c$ are odd. If that's the case, then the only prime value any of $p,q,r$ can take is $2$, for which the solution is $a=b=c=1$

QED.

Answer 3

Consider the case where(a=b=c=0). In this scenario:

(p = 0^0 + 0 = 1) (q = 0^0 + 0 = 1) (r = 0^0 + 0 = 1)

Since all three expressions yield the same result when (a=b=c=0), our assumption that all three are distinct is false. Therefore, at least two of the expressions must be the same.

Answer 4

Using parity, if $q,r$ are both odd,

Then $a $ odd $\implies c$ even $\implies b $ odd $\implies b^c$ odd $\implies p$ even.

Also, $a$ even $\implies c$ odd $\implies b $ even $\implies b^c$ even $\implies p$ even.

So let $p = 2 = b^{c} + a \implies a = 1 \implies b = 1 \implies q = 1 + c$ and $r = 1 + c \implies q = r$.

If one of $q,r$ is even, then WLOG let $q = 2 \implies p = r$ from above.