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

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Rajesh

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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 & \(\displaystyle n>1\)?
 
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Rajesh said:
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 & \(\displaystyle n>1\)?
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Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem
 
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.

[math] 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 [/math]
Hmm.... [imath]3^3 = 3^2*3[/imath]. That’s interesting. Can I generalize?

[math] 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).[/math]
Can you prove that that integer is not a perfect power?

Proofs do not leap off the page. Try things.
 
Rajesh said:
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 & \(\displaystyle n>1\)?
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Look at both sides modulo 4.
 
Rajesh said:
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 & \(\displaystyle n>1\)?
Click to expand...
4 days gone by - no peep from OP.
 
NotJeffM said:
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.
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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.
 
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