# Help needed with a proof

1. Oct 1, 2006

### 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?

Last edited: Oct 1, 2006
2. Oct 2, 2006

### Euclid

your proof seems correct to me

3. Oct 2, 2006

### HallsofIvy

Staff Emeritus
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.

4. Oct 2, 2006

### 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.

5. Oct 2, 2006

### 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 ?

6. Oct 2, 2006

### matt grime

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

7. Oct 2, 2006

### roger

less than b

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

8. Oct 2, 2006

### 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.

9. Oct 2, 2006

### matt grime

Now do you see how the above conlcusion was reached?

10. Oct 2, 2006

### 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

11. Oct 2, 2006

### roger

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

12. Oct 2, 2006

### shmoe

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

13. Oct 2, 2006

### murshid_islam

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".

Last edited: Oct 2, 2006
14. Oct 2, 2006

### HallsofIvy

Staff Emeritus
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!

15. Oct 2, 2006

### shmoe

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.

16. Oct 2, 2006

### murshid_islam

sorry, my mistake. you are absolutely right.

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