Squared Consecutive Integers

$a^{2} + (a + 3)^{2} = (a + 1)^{2} + (a + 2)^{2}$

$a^{2} + a^{2} + 6a + 9 = a^{2} + 2a + 1 + a^{2} + 4a + 4$

$6a + 9 = 6a + 5$

$9 = 5$

Therefore no solution

Since they are consecutive integers, then $a-b = c-d = -1$ . We have

$\begin{aligned} a^2 + d^2 &=& b^2 + c^2 \\ a^2 - b^2 &=& c^2 - d^2 \\ (a-b)(a+b)&=& (c-d)(c+d) \\ \xcancel{(a-b)}(a+b)&=& \xcancel{(c-d)}(c+d) \\ \end{aligned}$

But since $a<b<c<d$ , then $a+b < c + d$ , a contradiction! Thus, there is $\boxed0$ solution.