Is $\frac{a^2+b^2}{2}=c^2$ possible?

Question

I am looking for an integer solution to the equation: $$\frac{a^2+b^2}{2}=c^2(a\neq b\neq c)$$ That is a square number that is the mean of two other square numbers, is this possible? And if so please can you give me an example?

Answer 1

We know that $(x-y)^2+(x+y)^2=2(x^2+y^2).$

So, if $a=x+y$ and $b=x-y$, then $c^2=x^2+y^2.$

Now see Formulas for generating Pythagorean triples.

Answer 2

for example;a=1,b=7,c=5 or a=2,b=14,c=10

Answer 3

This means $a^2,c^2,b^2$ are in arithmetic progression. Without loss of generality assuming $a,b,c$ are mutually relatively prime, we have that $$a = p-q, b=p+q \text{ and }c = r$$ where $p$,$q$ and $r$ form Pythagorean triplets. In fact, all such squares in arithmetic progression must necessarily be of the form.

A detailed characterization is presented here.

Also, as an aside no four squares can be in arithmetic progression.