Perfect Squares: relatively prime and increasing arithmetic progression The squares 1, 25, 49 are pairwise relatively prime and in an increasing arithmetic progression. Let $x_1^2, y_1^2, z_1^2$ and $x_2^2, y_2^2, z_2^2$ be the next two triples with this property (so that 49, $z_1^2$, $z_2^2$ are the three smallest possible values of the third term). Find $z_1 + z_2$.
I really help about how to start/a good solution for this problem because I understand what the problem is asking but I'm not sure how to start. Thanks!
 A: First solution
Remark(1): Suppose that we can write the integer $r$ in the form $2Y^2-X^2$;
then we can conclude that each prime factor of $r$; 
which is of the type $8k \pm 3$,
must divides $r$, by an even power. 
In other words if $p \overset{8}{\equiv} \pm 3$ and 
$p^ \alpha \mid r$ and $p^ {\alpha+1} \nmid r$, 
then there is an integer $\beta$ such that: $\alpha=2\beta$. 
Moreover in this situation every solution of $2Y^2-X^2=r$ has the form: 
$X=p^ \beta X^ {\prime}$, $Y=p^ \beta Y^ {\prime}$ 
for some integers $X^{\prime}$ and $Y^{\prime}$.
List of the prime numbers of the form $8k + 1$ are as follows: 
$$17, 41, 73, 89, 97, ... \ \ \ ,$$ 
and the list of the prime numbers of the form $8k + 7$ are as follows: 
$$7, 23, 31, 47, 71, 79, ... \ \ \ .$$ 


Without loss of generality we can assume that: 
$x_i^2 < y_i^2 < z_i^2$, 
so we must have $x_i^2 + z_i^2 = 2y_i^2$; 
so we have $z_i^2 = 2y_i^2 -x_i^2$ and $x_i^2 = 2y_i^2 -z_i^2$. 
The minimum positive possible values for $(x_i,z_i)$ is equal to
$$(1,7); \ (7,17); \ (7,23); \ ... ,  $$
so $z_0=7$, $z_1=17$, $z_2=23$. 




Second solution
Remark(2): 
Let's remember that a parametric solution 
to the equation $x^2+y^2=z^2$ is given by: 
$$ 
             \left\{ \begin{array}{lcc}
             x=d(m^2-n^2) \ , 
             \\ 
             y=d(2mn) \ , 
             & 
             \\
             z=d(m^2+n^2) \ , 
             \\ 
             \end{array}
   \right.$$
where $m$ and $n$ are relatively prime integers with different pairity.


Remark(3): 
Now lets consider the equation $x^2+y^2=2z^2$; 
multiplying both sides by $2$ turns out: 
$$ 
(x+y)^2+(x-y)^2= 
2(x^2+y^2)= 
2(2z^2)= 
(2z)^2 . $$ 
Leting $x+y=d^{\prime}(m^2-n^2)$, 
$x-y=d^{\prime}(2mn)$ and $2z=d^{\prime}(m^2+n^2)$; 
and by a simple concusion 
we obtain the following parametric solution 
to the equation $x^2+y^2=2z^2$: 
$$ 
%%             \left\{ \begin{array}{lcc}
%%             x= \dfrac  {d         (m^2-n^2+2mn)         }{2} 
%%              = \dfrac  {d  \big(  (m+n)^2-2n^2  \big)  }{2} \ , 
%%             \\ 
%%             y= \dfrac  {d         (m^2-n^2-2mn)}{2} 
%%              = \dfrac  {d  \big(  (m-n)^2-2n^2  \big)  }{2} \ , 
%%             & 
%%             \\
%%             z= \dfrac  {d(m^2+n^2)}{2} \ , 
%%             \\ 
%%             \end{array}
%%   \right.$$
$$ 
             \left\{ \begin{array}{lcc}
             x= d         (m^2-n^2+2mn)          
              = d  \big(  (m+n)^2-2n^2  \big)   \ , 
             \\ 
             y= d         (m^2-n^2-2mn) 
              = d  \big(  (m-n)^2-2n^2  \big)   \ , 
             & 
             \\
             z= d(m^2+n^2) \ , 
             \\ 
             \end{array}
   \right.$$
where $m$ and $n$ are relatively prime integers with different pairity. 
So your problem reduces 
to minimizing the quantity 
$$\max \{ \ (m+n)^2-2n^2 \ , \ \ (m-n)^2-2n^2 \ \} . $$
after a simple computation you can get: 
$$(m, n)=(2, 1) \rightarrow (1, 7; 5), $$ 
$$(m, n)=(3, 2) \rightarrow (7,17;13), $$ 
$$(m, n)=(4, 1) \rightarrow (7,23;17). $$ 


Third solution: By using the primitivity assumption, 
we can conclude by an easy modular arithmetic module $8$,
that $r \overset{8}{\equiv} \pm 1$, 
so it only suffices to search thorough the integers $x_i$ and $z_i$ among the following list:
$$1,  9, 17, 25, ..., $$
$$7, 15, 23, ..., .$$
[For example note that $15$ is immpossible by lemma(1).]
A: I noted that each triplet begins with the last number of the previous triplet
For instance $(1,5,7)$ brings to $(7,13,17)$ and then to $(17,25,31)$
Going on for a while I figured out a formula that gives all this kind of triplets:
$$\left (2 n^2-1,2 n^2+2 n+1,2 n^2+4 n+1\right)$$
Indeed 
$$(2 n^2+2 n+1)^2-(2n^2-1)^2=8 n^3+12 n^2+4 n$$
and
$$(2n^2+4 n+1)^2-(2 n^2+2 n+1)^2=8 n^3+12 n^2+4 n$$
so squares  are in arithmetic progression.
The next triplets of this kind are $(7,\;13,\;17)$ and $(17,\;25,\;31)$
A: Famke's answer solves the problem. Still, there is indeed a direct correspondence between Pythagorean triples and your squares in arithmetic progression. [You asked about this in the original question, then alas you edited it out.]
Squares in arithmetic progression correspond to triples $d<e<f$ such that $f^2+d^2=2e^2$. And such triples correspond directly to Pythagorean triples, as follows.
a) If $f^2+d^2=2e^2$ for $d<e<f$, then $$e^2=(\frac{f-d}{2})^2 + (\frac{f+d}{2})^2$$ and we see that $\frac{f-d}{2}, \frac{f+d}{2}, e$ is a Pythagorean triple. 
b) If $a^2+b^2=c^2$ for $a<b<c$ (a Pythagorean triple), then $$(b+a)^2 + (b-a)^2 = 2c^2$$
and we see that $(b-a)^2, c^2, (b+a)^2$ are squares in arithmetic progression.
So another simpler way to solve your problem is to look at the smallest three primitive Pythagorean triples $(3,4,5),(5,12,13), (8,15,17)$. These yield the required $z_0=3+4=7, z_1=5+12=17, z_2=8+15=23$. 
