prove that there are infinitely many primitive pythagorean triples $x,y,z$ such that $y=x+1$ Here is my question: Prove that there are infinitely many primitive pythagorean triples $x,y,z$  such that $y=x+1$.
They gave me a hint that I should consider the triple $3x+2z+1, 3x+2z+2, 4x+3z+2$, but I honestly do not know how to start. 
Do I need contradiction?
 A: Start with the hint . . .

Let $a,b,c$ be given by
\begin{align*}
a &= 3x+2z+1\\[4pt]
b &= 3x+2z+2\\[4pt]
c &= 4x+3z+2
\end{align*}
where $x,z$ are unknown positive integers.

If we were lucky, the equation $a^2 + b^2 = c^2$ would hold identically, for all $x,z$. 

Let's try . . .
\begin{align*}
a^2 + b^2 - c^2 &= (3x+2z+1)^2 + (3x+2z+2)^2 - (4x+3z+2)^2\\[4pt]
&= 2x^2 + 2x + 1 - z^2\qquad\text{[by expanding and then combining like terms]}\\[4pt]
\end{align*}
So unfortunately, we were not that lucky, since $a^2 + b^2 - c^2$ simplifies to $2x^2 + 2x + 1 - z^2$, which is not identically zero. 

But what if it was the case that $2x^2 + 2x + 1 - z^2 = 0\,$ for some $x,z$? 

With that idea, let's look more closely at the equation $2x^2 + 2x + 1 - z^2 = 0$, to see if we can find a way to make it happen.

The key observation is 
\begin{align*}
&2x^2 + 2x + 1 - z^2 = 0\\[4pt]
\iff\; &2x^2 + 2x + 1 = z^2\\[4pt]
\iff\; &x^2 + (x^2 + 2x + 1) = z^2\\[4pt]
\iff\; &x^2 + (x+1)^2 = z^2\\[4pt]
\end{align*}
Given that all transitions above are of the form "if and only if", it follows that

$$(x,x+1,z)$$
$$\text{is a pythagorean triple}$$
$$\text{if and only if}$$
$$(3x+2z+1,3x+2z+2,4x+3z+2)$$
$$\text{is a pythagorean triple}$$

Thus, starting with any pythagorean triple of the form $(x,x+1,z)$ we can get a new one which is strictly larger (e.g., larger perimeter).

Note also that any pythagorean triple of the form $(x,x+1,z)$ is automatically a primitive pythagorean triple since $\gcd(x,x+1) = 1$.

Thus, all we need is one such triple to get started. Fortunately, that's easy$\,{\large{-}}\,$just start with the triple $(3,4,5)$. 

So we weren't so unlucky after all!
A: Their idea is: $x = 3a+2b+1, y = 3a+2b+2, z = 4a+3b+2$, and check that $x^2+y^2 = z^2$, and $(x,y,z) = 1$, with $a,b \in \mathbb{Z}$
A: $x^2 + y^2 = z^2$ and $y = x + 1 $   $\implies   (3x+2z+1)^2 + (3x+2z+2)^2 = (4x+3z+2)^2$
And this series goes on forever by $\cases{x := 3x+2z+1 \\
y := 3x+2z+2 \\
z := 4x+3z+2}$
$\ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \  $ (here $:=$ stands for assignment)
It is not very difficult to check that $\gcd(x,y,z) = 1$ at each step, thus ensuring the primitivity of the triplets.  
EDIT:
An example as to how a triplet is generated from the triplet in the previous step.
Suppose we start off from the triplet $3,4,5$ as indeed $4 = 3 + 1$.
Then the next triplet would be $(3 \times 3 + 2 \times 5 + 1, 3 \times 3 + 2 \times 5 + 2, 4 \times 3 + 3 \times 5 + 2) = (20,21,29)$.
And at the next step $(x,y,z) = (20,21,29)$.  
Hence the next triplet now would be $(3 \times 20 + 2 \times 29 + 1, 3 \times 20 + 2 \times 29 + 2, 4 \times 20 + 3 \times 29 + 2) = (119,120,169)$.
And at the next step $(x,y,z) = (119,120,169)$.
And this goes on, if you will, forever. 
A: Given the odd number $x$,
$$x^2=2y^2-1$$
which is just a Pell equation, then we get the Pythagorean triple,
$$\Big(\frac{x-1}2\Big)^2+\Big(\frac{x+1}2\Big)^2=y^2$$
The difference $d$ between the addends, of course, is $d=1$. This yields,
$$3^2+4^2=5^2\\20^2+21^2=29^2\\119^2+120^2=169^2$$
and so on.

P.S. If interested, a nice infinite counterpart is $a^2+b^2=(b+1)^2$ with solution,
$$(2m+1)^2+(2m^2+2m)^2 = (2m^2+2m+1)^2$$
which yields,
$$3^2+4^2=5^2\\5^2+12^2=13^2\\7^2+24^2=25^2$$
ad infinitum.
A: You may just ignore the hint and prove that $x^2+(x+1)^2=y^2$ has an infinite number of solutions in $\mathbb{N}\times\mathbb{N}$. Such identity is equivalent to
$$ (2x+1)^2-2 y^2 = -1 $$
hence it is enough to show that there are infinite natural solutions to $a^2-2b^2=-1$ with $a$ being odd. But, wait, if $a,b\in\mathbb{Z}$ and $a^2-2b^2=-1$ then $a$ has to be odd. And $a^2-2b^2$ is just the norm over $\mathbb{Z}[\sqrt{2}]$, where $3\pm2\sqrt{2}$ is an invertible element (unit norm). It follows that by raising $(3+2\sqrt{2})$ to some power we still get an invertible element, and a solution of $a^2-2b^2=1$. Since the norm of $1+\sqrt{2}$ is $-1$, in such a case $(a+2b)^2-2(a+b)^2=-1$ and we get a solution of $x^2+(x+1)^2=y^2$. For instance,


*

*$(3+2\sqrt{2})^3 = 99+70\sqrt{2}$

*$99^2-2\cdot 70^2 = 1$

*$239^2-2\cdot 169^2 = -1$

*$119^2+120^2=169^2$


or


*

*$(3+2\sqrt{2})^4 = 577+408\sqrt{2}$

*$577^2-2\cdot 408^2 = 1$

*$1393^2-2\cdot 985^2 = -1$

*$696^2+697^2=985^2$


In particular, if we define the Pell number $P_n$ as $\frac{(1+\sqrt{2})^n-(1-\sqrt{2})^n}{2\sqrt{2}}$, we have that

$$ \left(\frac{P_{2n-1}+3 P_{2n}-1}{2}\right)^2+\left(\frac{P_{2n-1}+3 P_{2n}+1}{2}\right)^2 = \left(P_{2n-1}+2\,P_{2n}\right)^2.$$

Such well-known identity can be simplified a bit.
A: By ignoring the hint and allowing $y^2 + x^2 = (x+1)^2 $ we get the following.
The difference between the square of any two adjacent numbers is equal to  the sum of those numbers;
$$ (x+1)^2 - x^2 = 2x + 1 = (x+1) + x $$
Every odd number will have an odd square that can be expressed as $2x + 1$.
Allowing $y^2 = 2x + 1$, we get
$$y^2 + x^2 = (x+1)^2 $$
Such a triple exists for every odd number.
