Is there a perfect square that is the sum of $3$ perfect squares? This is part of a bigger question, but it boils down to:
Is there a square number that is equal to the sum of three different square numbers?
I could only find a special case where two of the three are equal? https://pir2.forumeiros.com/t86615-soma-de-tres-quadrados (in portuguese).
Any clue?
 A: It really is a four variable parametrization, coming from quaternion multiplication. The attribution in wikipedia is to the number theorist V. A. Lebesgue. He did publish on this in the 1850's. However, this was surely known to Euler.
Discussed in detail in Pall 1940
I used those techniques, quaternions,  for related problems Find three numbers such that the sum of all three is a square and the sum of any two is a square and 
http://math.stackexchange.com/questions/1964607/when-will-a-parametric-solution-generate-all-possible-solutions/1965805#1965805  I was quite proud of this one, it gives a complete integer parametrization for $a^2 + b^2 + c^2 = 3 d^2$ using material from Jones and Pall 1939
Meanwhile, the first acceptable proof that ALL quadruples arise this way (primitive ones, that is) was due to L. E. Dickson in 1920.   There is a very nice article by Spira, let me find a link. I also have a pdf of Spira (1962).


A: There are many of them.
$$113^2 = 112^2+12^2 +9^2 $$
is an example. 
A: Dickson's History of the Theory of Numbers,
volume 2,
starting on page 261
has a number of solutions.
Here is one due to Euler:
$(p^2 + 1)^2(q^2 + 1)^2 
= (q^2 - 1)^2(p^2 + 1)^2 + 4q^2(p^2 - 1)^2 + (4pq)^2
$.
A: $$3^2+4^2+12^2=13^2{}{}{}{}{}{}{}{}$$
A: We know that $(n+1)^2-n^2=2n+1$, so pick your favorite Pythagorean triple $a^2+b^2=c^2$ with $c$ odd.  Let $c^2=2n+1, n=\frac {c^2-1}2$ and 
$$a^2+b^2+n^2=(n+1)^2$$
If you pick a triple with $c$ even, we can use $(n+2)^2-n^2=4n+4$, so we can let $n=\frac {c^2-4}4$ and have $$a^2+b^2+n^2=(n+2)^2$$
If $c$ is even, $c^2$ is divisible by $4$, so the division will come out even.
A: There are infinitely many: for any positive integer $n$ we have
$$n^2(n+1)^2+n^2+(n+1)^2=(n(n+1)+1)^2$$
