# [Number Theory] Prove (x^2 - y^2) is not equal to 6.

1. Aug 30, 2011

### razefast

1. The problem statement, all variables and given/known data
Prove the following proposition: For any positive integers x and y, (x^2 - y^2) is not equal to 6.

2. Relevant equations

3. The attempt at a solution

I'll try to prove using contradition.
Assume x^2 - y^2 = 6.
(x+y)(x-y) = 6

(x+y)=6 and (x-y)=1 (OR)
(x+y)=1 and (x-y)=6 (OR)
(x+y)=-6 and (x-y)=-1 (OR)
(x+y)=-1 and (x-y)=-6 (OR)
(x+y)=2 and (x-y)=3 (OR)
(x+y)=3 and (x-y)=2 (OR)
(x+y)=-2 and (x-y)=-3 (OR)
(x+y)=-3 and (x-y)=-2

When I solve those equations, x and y turns out to be fractions.
Therefore, (x+y)(x-y) is not equal to 6.
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Hi fellow physicsforumers
This is my first post. :D
I just started my discrete maths course at uni.
Is my solution correct?
Even if it is, is there any simpler way?
Thanks,
razefast

2. Aug 30, 2011

### lanedance

so its a proof of contradiction using cases - looks reasonable to me, though haven't done a heap of number theory

i think though you need to state somewhere that the only 2 integer factorisations of 6 are
1.6
2.3
-2.-3
-1.-6

interestingly, the brute force method is almost quicker here
clearly x>y>0 for this to be true
so say
x=2, y=1, x^2-y^2=3
x=3, y=2, x^2-y^2=5
x=3, y=1, x^2-y^2=8
x=4, y=3, x^2-y^2=7

however you would also need to show x^2-(x-1)^2 is an monotonically increasing function, but that shouldn't be too hard

3. Aug 30, 2011

### razefast

Hi lanedance,

Actually, I have no idea how to prove x^2-(x-1)^2 is an monotonically increasing function :(
I have never learnt about monotonic function so I had to google it but still confused. lol

With regards,
razefast

4. Aug 30, 2011

### process91

x^2 - (x-1)^2 is easy to prove as monotonic, even without calculus. I will give a different example so you get the idea.

Monotonically Increasing Proof:
f(x) = x^2+3x+1
Prove that f(x) is monotonically increasing for x>0, that is,
If 0<a<b, then f(a)<f(b)

Let d=b-a => b=a+d.
f(b)=f(a+d)=(a+d)^2 + 3(a+d) +1 = a^2 + 2ad + d^2 + 3a + 3d + 1 = f(a) + (2a + d + 3)d
Now since 0<a<b, b-a>0 and so d>0. Thus (2a + d + 3)d > 0, and so f(b)>f(a).

The problem of proving x^2 - (x-1)^2 is monotonic should be similar. In fact, it is a little algebraically simpler, and you should not need a restriction of x>0.

5. Aug 30, 2011

### lanedance

monotonically increasing means it is always increasing as x gets bigger, or the derivative is always greater than zero

in terms of the proof, we just want to show the gap between x^2 and (x-1)^2 is always greater than 6 for x>4. That way we don't have to check any values of x>4

now expanding the difference
x^2-(x-1)^2 = x^2-(x^2-2x+1) = 2x-1

which is a straight line, hopefully you can convince yourself this is always increasing.

Now say you have two integers s and t, if t>s can you show 2t+1>2s+1? That would be enough.

All that said your original proof was fine with the extra line
. This is just another way of looking at it.

6. Aug 31, 2011

### razefast

Hi lanedance and process91

Those explanations are quite clear.
I learnt new things today.
Thx for helping out de newbie xD
Man..Physics forums is great ^^

7. Aug 31, 2011

### verty

Here's another way, let x = y + k and try to derive a contradiction.