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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!

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    $\begingroup$ Write a system of equations. So there was no confusion in the formulation of the problem. $\endgroup$ – individ Aug 4 '17 at 13:32
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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).]

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  • $\begingroup$ Thank you so much! This was really helpful and I think I understand how to do the problem now. $\endgroup$ – Annie Smith Aug 4 '17 at 16:27
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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 $(17,\;25,\;31)$

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  • $\begingroup$ There is an smaller solution other than $(17,25,31)$; namely $(7,17,23)$, with the sum $23+17=40$. $\endgroup$ – Davood Khajehpour Aug 4 '17 at 17:48
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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 $z_0=3+4=7, z_1=5+12=17, z_2=8+15=23$.

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