You are Here: Home >< Maths

# Help with this proof. Watch

1. How do you go about proving that "1 + 10^n" cannot be a square number.

I did this:

1 + 10^n = a^2
10^n = a^2 - 1
10^n = (a-1)(a+1)

then nlog10 = log(a-1) +log(a+1)

and since nlog10 is an integer then log(a-1) + log(a+1) must be an integer. As a must be an integer then there is no value of a that gives both (a-1) and (a+1) as bases of 10.
Hence 1 + 10^n cannot be a square number.

I'm not sure if this is an accepted proof or not . Can't find anything from google. How would you guys go about doing it??

Cheers
2. (Original post by jaheen22)

then nlog10 = log(a-1) +log(a+1)

and since nlog10 is an integer then log(a-1) + log(a+1) must be an integer. As a must be an integer then there is no value of a that gives both (a-1) and (a+1) as bases of 10.
Can you explain this a bit more?
3. You can do this using mods.

Under mod 9

The remainder when a number is divided by 9 is the same as the remainder of the sum of the digits of that number. The sum of the digits of A is always 2.

11
101
1001
10001
100001
... etc

Now if you investigate the square numbers in mod 9.

But you can't make 2 mod 9 using a square number in mod 9, so A will never be a square number.
4. (Original post by Daniel Freedman)
Can you explain this a bit more?
I've thought of it differently. Seeing as we're only dealing with positive integers then the only factors of 10 are 1,2,5 and 10.

now: nlog10 = n(log 2 +log 5) or n(log 1 + log10)

log(a-1) + log(a+1) = n(log 2 + log5)

I don't know if that would help my arguement though. I don't know what to do anymore
5. (Original post by Noble.)
You can do this using mods.

Under mod 9

The remainder when a number is divided by 9 is the same as the remainder of the sum of the digits of that number. The sum of the digits of A is always 2.

11
101
1001
10001
100001
... etc

Now if you investigate the square numbers in mod 9.

But you can't make 2 mod 9 using a square number in mod 9, so A will never be a square number.
Ah, I see, I wouldn't have considered modular arithmetic before because we haven't touched upon it yet in maths. Thanks mate

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: December 7, 2010
Today on TSR

### Degrees to get rich!

... and the ones that won't

### Women equal with Men?

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.