# Number Theory ProblemWatch

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#1
Given that a, b and n are positive integers, and

(a^2 + b^2)/(ab - 1) = n

Prove that n = 5

First one who proves it gets rep.

Galois.
0
14 years ago
#2
Are you sure it's n=5? For some reason I got n=4...

I gotta go, so I'll check my steps to see if I added wrong or something when I get back.
0
#3
(Original post by dvs)
Are you sure it's n=5? For some reason I got n=4...

I gotta go, so I'll check my steps to see if I added wrong or something when I get back.
No n = 5 is the only integer solution to that expression (apparently).

What values of a and b did you use?

Galois.
0
14 years ago
#4
n = 5 is OK

a = 1 and b = 2 or vice versa

the value of the LHS decreases as a and b increase, so the above is the only solution, but I have no idea about a proof!

Aitch
0
14 years ago
#5
(Original post by Aitch)
n = 5 is OK

a = 1 and b = 2 or vice versa

the value of the LHS decreases as a and b increase, so the above is the only solution, but I have no idea about a proof!

Aitch
Hmm a=1 and b = 3
0
14 years ago
#6
(Original post by RichE)
Hmm a=1 and b = 3
You're right!

The other half of what I wrote is wrong too!

Aitch
0
14 years ago
#7
(Original post by Aitch)
You're right!

The other half of what I wrote is wrong too!

Aitch
...but not the bit about not having any idea about a proof. That bit is entirely accurate.

Aitch
0
#8
a=2 and b=1 (and vice versa)
a=3 and b=1 ""
a=9 and b=2 ""
a=14 and b=3 ""
a=43 and b=9 ""
a=67 and b=14 ""

There are no more pairs with a and b both less than 100. There are an infinite number of pairs a and b, but the question isn't find a and b, the question is prove n = 5.

Any one?

Galois.
0
14 years ago
#9
I've not done it, but I got another one for number theory.
a>b>c>d are positive integers
ac+bd can divide by (a+b+d-c).
Prove that ab+cd is not a prime number.
0
14 years ago
#10
Suppose that it is prime, then a+b+d-c must be that same prime and equal to ab+cd;
a+b+d-c = ab + cd
a(b-1) + d(c-1) = b-c
c > b so the RHS is negative, whereas the LHS must be positive. So we have a contradiction.
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#11
(Original post by JamesF)
Suppose that it is prime, then a+b+d-c must be that same prime and equal to ab+cd;
a+b+d-c = ab + cd
a(b-1) + d(c-1) = b-c
c > b so the RHS is negative, whereas the LHS must be positive. So we have a contradiction.
Who said c was greater than b? The RHS is also positive, because b > c.

Galois.
0
14 years ago
#12
(Original post by Galois)
Who said c was greater than b? The RHS is also positive, because b > c.

Galois.
Crap, read the inequality signs the wrong way round
0
#13
Can any one do the first question?

I am calling for the best number theorist here.

Galois.
0
14 years ago
#14
God, this is so frustrating. I must do it.
0
14 years ago
#15
Ahrr, it made my brain exploding. Dang! I have alot of this type of questions and never end up with successful answers. I'll post it after someone do Galois's question.
0
#16
(Original post by J.F.N)
God, this is so frustrating. I must do it.
Any luck?
0
14 years ago
#17
I will do it. By tonight. Don't give the answer. I must do it.
0
#18
(Original post by J.F.N)
I will do it. By tonight. Don't give the answer. I must do it.
OK. Remember solution in white please.

Galois.
0
14 years ago
#19
What do you mean, in white?
0
14 years ago
#20
(Original post by Galois)
OK. Remember solution in white please.

Galois.
What sort of level is this? I'm not intending to stay awake all night to do this... but...

It looks interesting, I've done P4, but can find no way into this.

Have tried a few approaches, but end up with too many variables.

When you put up the answer, and put the contenders out of their misery, can you put up a reading list, or similar?

Thanks.

Aitch
0
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