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Are there any 3 natural numbers that satisfy $a^2+b^2=2z^2 $?

This is a question that arised as I was trying to solve another question: Is there an arithmetic progression, of natural numbers in which three (not necessarily successors) elements perform a geometric progression?

Thank you!

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    $\begingroup$ Yes, try $a=b=z=1$. $\endgroup$
    – vadim123
    Jun 29, 2014 at 15:24
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    $\begingroup$ Or $a= b=z = k$ for any $k$ $\endgroup$
    – Mathmo123
    Jun 29, 2014 at 15:26
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    $\begingroup$ For a non-trivial answer, $1^2+7^2=2\cdot5^2$ $\endgroup$
    – G Tony Jacobs
    Jun 29, 2014 at 15:31

2 Answers 2

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recognize the identity:

$$(a-b)^2+(a+b)^2=2(a^2+b^2)$$

If you let $x = a - b$, and $y = a + b$, and $z = (a^2+b^2)$

we then have:

$$ x^2 + y^2 = 2z^2$$

Which connects back to your equation.

EDIT

In response to your other question, there can exist 3 natural numbers for which it can form both a arithmetic and geometric sequence (not in succesion).

i.e. $2 + 4 + 6 + .....$ Arithmetic Sequence, Common Difference: 2, First term: 2

$2 + 4 + 8 + 16 + .....$ Geometric Sequence, Common Ratio: 2, First term: 2

As you can see, the first 3 terms of this Geometric sequence are also part of an arithmetic sequence (again, not in succession).

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$$(p-q)^2+(p+q)^2=2(p^2+q^2)$$

Now we need $\displaystyle p^2+q^2=r^2$ set $\displaystyle p=s^2-t^2,q=2st$

Reference : Formulas for generating Pythagorean triples

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