Never be a perfect power: Prove 11a^(2m) + 18a + 25 = z^n is impossible, for a an odd integer, m >= 0, n > 1

[Rajesh]

How do you show that it's Impossible to write $\displaystyle 11a^{2m}+18a+25=z^n$, where $\displaystyle a$ is any positive odd integer, $\displaystyle m$ is any positive integer & $\displaystyle n>1$?

 

[khansaheb]

How do you show that it's Impossible to write $\displaystyle 11a^{2m}+18a+25=z^n$, where $\displaystyle a$ is any positive odd integer, $\displaystyle m$ is any positive integer & $\displaystyle n>1$?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

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Please share your work/thoughts about this problem

 

[NotJeffM]

I can think of at least three strategies for a proof: double weak induction, factoring, or contradiction. But the first thing I would do is to experiment.

$$11(2 * 1 - 1)^{2*0} + 18(2*1-1) + 25 = 11*1+18*1+25= 11+18+25=54 = 2*3^3.\\ 11(2*2-1)^{2*0}+16(2*2-1)+25=11*1+18*3+25= 54+36=90=2*3^2*5.\\ 11(2*3-1)^{2*0}+18(2*3-1)+25= 11*1+18*5+25=90+36=126=2*3^2*7$$
Hmm.... $3^3 = 3^2*3$. That’s interesting. Can I generalize?

$$11(2j-1)^{2*0}+18(2j-1)+25=11*1+36j-18+25=36j+36-18=18(2j+1)=\\ 2*3^2*(2j+1).$$
Can you prove that that integer is not a perfect power?

Proofs do not leap off the page. Try things.

 

[blamocur]

How do you show that it's Impossible to write $\displaystyle 11a^{2m}+18a+25=z^n$, where $\displaystyle a$ is any positive odd integer, $\displaystyle m$ is any positive integer & $\displaystyle n>1$?

Look at both sides modulo 4.

 

[khansaheb]

How do you show that it's Impossible to write $\displaystyle 11a^{2m}+18a+25=z^n$, where $\displaystyle a$ is any positive odd integer, $\displaystyle m$ is any positive integer & $\displaystyle n>1$?

4 days gone by - no peep from OP.

 

[NotJeffM]

4 days gone by - no peep from OP.

I think these threads that die at birth are either (1) I came here to avoid work, fools, or (2) now that I asked the question, the answer is obvious to me. VERY different psychological processes leading to the same result.

 

[Steven G]

Let's not call students fools.

 

[blamocur]

I think these threads that die at birth are either (1) I came here to avoid work, fools, or (2) now that I asked the question, the answer is obvious to me. VERY different psychological processes leading to the same result.

Occasionally I give likes to students whose reaction I like. For example, I like when students respond at the end instead of simply disappearing and leaving us wondering whether our effort to help was of any use at all.

 

[NotJeffM]

Let's not call students fools.

I wasn’t. I was speaking in the voice of those OP’s who come here expecting that we will do their homework for them.

 

[Steven G]

I wasn’t. I was speaking in the voice of those OP’s who come here expecting that we will do their homework for them.

OK!