This is the first post of a series devoted to Number theory. I decided to start it off with a well-known fact about the primes in the form \(4k+3\).This fact can be extremely useful in solving problems.

**Lemma**: *If \(x^{2}+y^{2}\) is divisible by a prime in the form \(p=4k+3\), then p\(\mid\)x and p\(\mid\)y.*

**Proof**:First let's note that if one of the numbers x and y is divisible by p, then so is the other.Let's prove the lemma by contradiction (*reductio ad absurdum*).
So let \(x^2+y^2\equiv0\pmod{p}\) and both x and y are coprime to p.Then \(x^2\equiv-y^2\pmod{p}\Rightarrow x^{p-1}\equiv-y^{p-1}\pmod{p}\).We have just risen the both sides to the power of *\(\frac{p-1}{2}=2k+1\)*.But by Fermat's little theorem (since x and y are coprime to p) we get that \(1\equiv-1\pmod{p}\), which is impossible \(\Rightarrow\)*p divides at least one of the numbers x and y , but as we said before, that means that p divides both of them!*

So from this lemma we can see that if a number can be represented as the sum of two squares of natural numbers, then in its prime factorization primes of the form *4k+3* are included in an even power. So if \(n = a^2 + b^2\), then \(\upsilon_p(n)=2r\) for some natural numbers a, b and r.Now for a little bit of exercise I will post a problem for you to try yourself.

**Problem**:Show that \(4xy -x-y=z^2\) has no solutions in natural numbers.

**Hint**:Try to factorize.

## Comments

Sort by:

TopNewestMultiplying by 4 and adding 1 yields:

\((4x-1)(4y-1)=4z^2+1\)

Now let \(p=4k+3\) be a prime for some integer \(k\). We have that \(p \mid LHS\) so \(p\mid RHS\). By the theorem in the post, we must then have \(p \mid 1\). But \(p>1\), so there are no solutions. – Daniel Remo · 3 years, 6 months ago

Log in to reply

– Eddie The Head · 3 years, 6 months ago

Since both the factors on the LHS are of the form 4k+3 it must have a prime fsctor of that form and hence your proof follows right?Log in to reply

– Bogdan Simeonov · 3 years, 6 months ago

Yep, that is how it's supposed to be.Log in to reply

I am sorry if the title was misleading!Anyways, if you found this post helpful in any way, make sure to keep track of my future posts!Also, make sure to post your solutions to the problem at the end of the post! – Bogdan Simeonov · 3 years, 6 months ago

Log in to reply

A good exercise is also by this lemma, and Fermat's theorem on sum of two squares, characterize all integers that can be represented as sum of two squares. – Yong See Foo · 3 years, 6 months ago

Log in to reply

– Bogdan Simeonov · 3 years, 6 months ago

Maybe I will do that...but I think it's better by starting with something simpler.Log in to reply

– Yong See Foo · 3 years, 6 months ago

Yea it is a nice result by this. That Fermat theorem might be pretty hard to prove...But maybe you can consider.Log in to reply

– Michael Tang · 3 years, 6 months ago

Fermat's result is very difficult to prove - it uses quadratic residues, order, and Gaussian integers, pulling together many of the important facts of elementary number theory.Log in to reply

– Yong See Foo · 3 years, 6 months ago

It is hard, but there are several proofs...Check out Euler's proof using infinite descent,which uses pretty elementary number theory. I also saw a proof that uses Thue's theorem.Log in to reply

– Bogdan Simeonov · 3 years, 6 months ago

The infinite descent proof was about Fermat's Last Theorem for the cubes, wasn't it?Or is there such a proof for the squares theorem?Log in to reply

– Yong See Foo · 3 years, 6 months ago

There is an infinite descent proof for this too.Log in to reply