# Thread: Two HARD algebra problems.

1. ## Two HARD algebra problems.

Problem 15
A pair (x,y) of positive integers is called square if x + y and xy are both perfect squares. For example, pair (5,20) is square since 5 + 20 = 52 and 5 x 20 = 102. Prove that no square pairs exists in which one of its numbers is 3.
Problem 16
Primes p and q are called twin primes, if q = p + 2. Prove that the numbers p^4 + 4 and q^4 + 4 are never relatively prime, if p and q are twin primes.

2. For 2), assume p,q are twin primes, that is $p = q + 2$.
Then, by the binomial theorem:
$p^4 = (q+2)^4 = \sum_{k=0}^4{\binom{4}{k}\cdot q^k \cdot 2^{n-k}}$

Now, since for each $1 \leq k \leq 4, 4 | \binom{4}{k}$, we get that $p^4 + 4 = q^4 + 4 + 4t, \text{ for some } t \in \mathbb{N}$

Can you finish from here?

3. Defunkt, how come you assumed that 3 + y = p^2 and 3y = q^2? In particular, why a singular prime? couldn't it be (p * q^3 * r^2)^2 for example?

4. Originally Posted by Bingk
Defunkt, how come you assumed that 3 + y = p^2 and 3y = q^2? In particular, why a singular prime? couldn't it be (p * q^3 * r^2)^2 for example?
Err, yes, morning brainfart my bad. I'll revise the proof.

5. Originally Posted by Defunkt
Err, yes, morning brainfart my bad. I'll revise the proof.
Plz see below

6. Originally Posted by Defunkt
Err, yes, morning brainfart my bad. I'll revise the proof.
This should work -

3+x = a^2
3x = b^2

so, 9 + b^2 = 3a^2

3|b^2 => 9|b^2 => 3|a^2 => 9|a^2

Divide the whole equation by 9. We get
1 + m^2 = 3n^2

Note: either of
m = 3k, 3k+1, 3k+2 doesn't satisfy the above.

7. Hehe, I like that term ... morning brainfart

aman_cc, how did you get that 3|b^2 => 9|b^2?

I got that also, but I'm just wondering if we did it the same way (I like how alot (in my limited experience) of things in number theory can be proven in different ways )

8. Is there any other method of doing this problem other than binomial theorem?

That looks way too advanced for me. I'm only in 10th grade.