<s>[INST] Show that there are infinitely many natural numbers such that $a^2+b^2=c^2+3 .$ This question was asked in the Crux Mathematicorum , October edition , Pg -$5$  , which can found be Here.
The question states that :

Both $4$ and $52$ can be expressed as the sum of two squares as well as
  exceeding another square by $3$ :
  $$4 = 0^2+2^2 \quad\,, \, 4-3=1^2 $$
$$52 = 4^2+6^2 \quad\,, \, 52-3=7^2 $$
  Show that there are an infinite number of such numbers that have these two
  characteristics.

My attempt :
I Found $4,52$ and $292$ to have this characteristics. An interesting feature I noticed was $$4 = \color{red}{0^2}+2^2 \quad\,, \, 4-3=\color{green}{1^2}$$
$$52 =4^2 + \color{red}{6^2} \quad\,, \, 52-3=\color{green}{7^2}$$
$$292 = 6^2 + \color{red}{16^2} \quad\,, \, 292-3=\color{green}{17^2}$$
If a number $y$ satisfies this property , then it might be written as :
$$y= a^2+b^2 = c^2+3$$
And based on the above examples , I conjectured that :
$$(k)^2 + b ^2 = (k+1)^2 + 3$$
where $a=k$ and $c=k+1$.
This expression on simplifying gives us :
$b^2 = 2(k+2)$ . For R.H.S to be a perfect square , $(k+2)$ must be of the form $2x^2$  , which on solving ,  gives $k = 2x^2-2$ and $b$ comes out to be $2x$.
So a solution is given by $\color{blue}{(a,b,c) = (2x^2-2,2x,2x^2-1)}$
And our number becomes $y = 4(x^4 - x^2 + 1)$ for all $x\in\mathbb{N}$ and hence there are infinite numbers with this characteristics. 

Although ' maybe ' this proves the question , it is not quite rigorous method to do this . Also , it does not provide all the possible numbers as we have only taken the special case when $a=k \,\,, c= k+1$. 
What is the better way to solve this problem and the general formula for the number?

Bonus Question : Prove that the highest power of two dividing the number is $2.$ Or more generally ,show that :
$$2^c\nmid y \quad \quad \text { For any } c \ge 3.$$
Verified it for all $y\le1.5\times10^5$ and it seems quite likely to be true . For my case , where $y = 4(x^4 - x^2 +1)$ , this is obvious as $x^4-x^2+1$ is always odd and hence the number is only divisible by $4$. But what about the other numbers ? 

Edit :
The first $5$ numbers with this characteristics are :
 $$4 = 0^2+2^2 \quad\,, \, 4-3={1^2}$$
$$52 =4^2 + {6^2} \quad\,, \, 52-3=7^2$$
$$292 = 6^2 + {16^2} \quad\,, \, 292-3={17^2}$$
$$628 = 12^2 + {22^2} \quad\,, \, 628-3={25^2}$$
$$964 = 8^2 + {30^2} \quad\,, \, 964-3={31^2}$$
 [/INST] Brainstorming from doing something similar with pythagorean triples.  
If $b^2 = 2a + 1 + 3=2a+4$ then $a^2 + b^2 = a^2 + 2a + 1 + 3 = (a+1)^2 + 3$.
If $b$ is any even number,  $2k; k > 1$  then if $a = \frac {(2k)^2-4}2= 2k^2-2$
And if $c = 2k^2 -1$ then
$a^2 + b^2 = $
$(2k^2 -2)^2 + 4k^2 =$
$4k^4 - 8k^2 + 4 + 4k^2 =$
$4k^4 - 4k^2 + 1 + 3 =$
$(2k^2 - 1)^2 + 3=$
$c^2 + 3$.
Yep.... that seems to do it.
=====
Perhaps more formally I should have done
$a^2 + b^2 = c^2 + 3 \implies$
$c^2 - a^2 = b^2 -3$
$(c-a)(c+a) = b^2 -3$
If I let $d = c-a$ be an odd number and $e = d+2a = c+a$ be a larger odd number then so long as we have
$b^2 -3 = d*e$ is an odd number then we have a solutions.
So if $b$ is any even number $> 2$ then $b^2 -3$ is odd number.  If we let $b^2 -3 = d*e; d< e$ be any factors of $b^2-3$ (if $b^2 -3$ is prime or a prime square we can let $d = 1$ and $e=b^2 -3$).  Then we let $a = \frac {e-d}2$ and $c = \frac {e+d}2$ we then have
$b^2 -3 = d*e = (c-a)(c+a) = c^2 - a^2$ and 
$a^2 +b^2 = c^2 + 3$.
...
If $b$ is odd and $b^2 -3$ is even then $b= 2k+1$ and $b^2 -3 = 4k^2 + 4k -2$ and is divisible by $2$ but not $4$.  $c-a$ and $c+a$ must both be the same parity so this will not be possible.
</s>