How many triangles?

For $(a, b, c)$ to be a triangle, we require that $a + b > c \\ b + c > a \\ c + a > b \\$ From the original equation, for $c$ to be positive, necessarily $a + b \leq 20$ . But if $a + b \leq 19$ , then $c >= 39$ , thereby contradicting $a + b > c$ . Therefore, since $a$ and $b$ are integers, we deduce $a + b = 20$ and $c = 15$ .

From this, $b + c > a \\ \Rightarrow 20 - a + 15 > a \\ \Rightarrow 35 > 2a \\ \Rightarrow a \leq 17 \\$ and similarly $c + a > b \\ \Rightarrow 15 + a > 20 - a \\ \Rightarrow 2a > 5 \\ \Rightarrow a \geq 3 \\$ In the same way, $3 \leq b \leq 17$ . The relation $a + b = 20$ puts these values of $a$ and $b$ in one-to-one correspondence with each other. Therefore, there are 15 possible solutions.

${ \quad \quad \quad \quad (a+b) }^{ 2 }+c=415\\ \Longrightarrow \quad c=415-{ (a+b) }^{ 2 }\\ So\quad it\quad is\quad clear\quad that\quad 0\le (a+b)\le 20\quad [as\quad c\quad is\quad a\quad positive\quad integer]\\ Case\quad A:\quad Now,if\quad (a+b)=20\quad then\quad c=415-{ 20 }^{ 2 }=15\\ For\quad forming\quad the\quad sides\quad of\quad a\quad non-degenerate\quad triangle,the\quad summation\quad of\quad any\quad two\quad numbers\quad of\quad the\quad triplet\quad (a,b,c)\quad must\quad be\quad greater\quad than\quad the\quad third\quad one.\\ Maintaining\quad this\quad condition,\quad the\quad correct\quad triplets\quad are\quad (3,17,15),(4,16,15),(5,15,15),(6,14,15),(7,13,15),(8,12,15),(9,11,15),(10,10,15),(11,9,15),(12,8,15),(13,7,15),(14,6,15),(15,5,15),(16,4,15),(17,3,15)\quad total\quad 15\quad triplets.\\ Case\quad B:\quad if\quad (a+b)=19,\quad c=415-{ 19 }^{ 2 }=54.\\ So,it\quad is\quad obvious\quad that\quad if\quad (a+b)\le 19,\quad c\ge 54\\ in\quad this\quad case\quad c\quad becomes\quad larger\quad than\quad (a+b),\quad and\quad so\quad no\quad non-degenerate\quad triangle\quad can\quad be\quad formed\quad having\quad sides\quad a,b,c.\\ So\quad the\quad number\quad of\quad distinct\quad ordered\quad triplets\quad is\quad \boxed{15}$