Proving n2 - 19n + 89 is Not a Perfect Square for n>11

[murshid_islam]

hi, i have to prove that if n is a perfect square and n>11, then n² - 19n + 89 is not a perfect square. i have came up with the following:

n>11
100 - 89 < 20n - 19n
-20n + 100 < -19n + 89
n² - 20n + 100 < n² - 19n + 89
(n-10)² < n² - 19n + 89......(1)

n>11
92 - 81 < 19n - 18n
-19n + 92 < -18n + 81
n² - 19n + 92 < n² - 18n + 81
n² - 19n + 89 + 3 < (n-9)²
n² - 19n + 89 < (n-9)²......(2)

combining (1) and (2), we get,
(n-10)² < n² - 19n + 89 < (n-9)²

since n² - 19n + 89 is between two consecutive perfect squares, it cannot be a perfect square itself. (QED)

but my proof doesn't require n to be a perfect square (as stated in the problem). is the question wrong? or am i making some mistake in my proof?

 

[Euclid]

your proof seems correct to me

 

[HallsofIvy]

Indeed, you don't need the n> 11 either. n²- 19n+ 89 is not a perfect square number for any n because n²- 19n+ 89 is not a perfect square polynomial.

 

[matt grime]

I'm not sure I get that, Halls.

x^2+9 is not a perfect square polynomial, yet when x=4, x^2+9=25. There are of course an infinite number of examples, I just wanted one where we evaluate x at a perfect square.

 

[roger]

I don't understand Halls either and I don't understand how the op got from
n^2 - 19n + 89 + 3 < (n-9)^2
to
n^2 - 19n + 89 < (n-9)^2 ?

 

[matt grime]

If a is less then b, is a-3 less than, more than, or equal to b?

 

[roger]

less than b

by the way, please could you let me have a look at the maths questions you used to have ?

 

[shmoe]

This polynomial is also a perfect square at n=11.

The "n a perfect square" isn't necessary with the n>11 condition. That doesn't make the question wrong, just uneccesarily weaker than it could have been. You could actually remove either condition (n a square or n>11) and it would still be correct.

 

[matt grime]

less than b

Now do you see how the above conlcusion was reached?

 

[roger]

but what did Halls mean by :n2- 19n+ 89 is not a perfect square number for any n because n2- 19n+ 89 is not a perfect square polynomial

 

[roger]

Now do you see how the above conlcusion was reached?

yes but how would it be proven explicitly without using proof by contradiction ?

 

[shmoe]

yes but how would it be proven explicitly without using proof by contradiction ?

You agree

n^2 - 19n + 89 < n^2 - 19n + 89 + 3

right? So

n^2 - 19n + 89 + 3 < (n-9)^2

implies

n^2 - 19n + 89 < (n-9)^2

"<" is transitive, a<b and b<c implies a<c

 

[murshid_islam]

You could actually remove either condition (n a square or n>11) and it would still be correct.

i think if we remove the condition "n>11", then it wouldn't be correct.
because, then n² - 19n + 89 can be perfect square for n = 11.
but we can remove the condition "n is a perfect square".

 

[HallsofIvy]

I don't understand it myself! Another case of shooting from the hip.

I'm tempted to go back and delete that post and pretend I never said any such thing!

 

[shmoe]

i think if we remove the condition "n>11", then it wouldn't be correct.
because, then n² - 19n + 89 can be perfect square for n = 11.

I said as much above, but notice n=11 is not a perfect square. I said "either" condition, not "both". Maybe I should have specified that explicitly.

 

[murshid_islam]

I said "either" condition, not "both".

sorry, my mistake. you are absolutely right.