jmc

Let $S=a_1+a_2+a_3+\cdots +a_{17}.$ Then from the given condition, $$a_i^2=S-a_i$$for all $1\le i\le 17.$ In other words, each $a_i$ is a root of $$x^2+x-S=0.$$This quadratic has at most two roots, which means that there are at most two different values among the $a_i,$ for any particular 17-tuple.

Suppose that all the $a_i$ are equal, say $$a=a_1=a_2=a_3=\cdots =a_{17}.$$Then $S=17a,$ so from the equation $x^2+x-S=0,$ $$a^2+a-17a=0.$$Then $a^2-16a=a(a-16)=0,$ so $a=0$ or $a=16.$

Otherwise, there are exactly two different values among the $a_i,$ say $a$ and $b.$ Suppose that $n$ of the $a_i$ are equal to $a,$ so the remaining $17-n$ values are equal to $b,$ where $1\le n\le 16.$ Then $$S=na+(17-n)b.$$Since $a$ and $b$ are the roots of $x^2+x-S=0,$ by Vieta's formulas, $a+b=-1$ and $ab=-S.$ Hence, $$na+(17-n)b=-ab.$$From $a+b=-1,$ $b=-a-1.$ Substituting, we get $$na+(17-n)(-a-1)=-a(-a-1).$$This simplifies to $$a^2+(-2n+18)a-n+17=0.(\ast )$$Since $a$ is an integer, the discriminant of this polynomial must be a perfect square. Thus, $$(-2n+18{)}^2-4(-n+17)=4n^2-68n+256=4(n^2-17n+64)$$is a perfect square, which means $n^2-17n+64$ is a perfect square.

Checking all values in $1\le a\le 16,$ we find that $n^2-17n+64$ is a perfect square only for $n=5$ and $n=12.$

For $n=5,$ equation $(\ast )$ becomes $$a^2+8a+12=(a+2)(a+6)=0,$$so $a=-2$ or $a=-6.$ The respective values of $b$ are $b=1$ and $b=5.$

So one possibility is that five of the $a_i$ are equal to $-2,$ and the remaining 12 are equal to 1. There are $\left(\genfrac{}{}{0ex}{}{17}{5}\right)=6188$ 17-tuples of this form. Another possibility is that five of the $a_i$ are equal to $-6,$ and the remaining 12 are equal to 5. There are $\left(\genfrac{}{}{0ex}{}{17}{5}\right)=6188$ 17-tuples of this form.

The case $n=12$ leads to the same possibilities. Therefore, the total number of 17-tuples is $2+6188+6188=\boxed{12378}.$