now you have to sing it again or turn into a bathtub
6.
With a TI “grapher”, I’ve just found that
244^2 == 5 (mod 1009). One simply puts
Y_1 = 1009(fPart(X^2/1009))…
this computes the equivalence-class-mod-1009
of the square of each input successively…
and “scrolls down” in the TABLE window
(set to display natural number X’s).
Hence x^2 == 5 (mod 1009) *does* have a solution.
(In fact, two of them; the other is of course
765 (= 1009 – 244)).
In detail,
244^2 = 58536 = 59*1009 + 5
and
765^2 = 585225 =580*1009 + 5.
(The point here is that one may easily
*verify* such results by “paper-&-pencil”
methods.)
To find the roots *entirely* by p-&-p methods,
the best line of attack would probably be via
“primitive roots”.
trouble is, is, that that “5”
was supposed to’ve been “150”.
once more, dear friends!
(or close the wall up with our
english dead!)
the livingston review 
April 11, 2014 at 11:17 am
3.2.6
With a TI “grapher”, I’ve just found that
139^2 == 150 (mod 1009).
One simply puts
Y_1 = 1009(fPart(X^2/1009))…
this computes equivalence-classes-mod-1009
of the squares of each of the inputs successively…
and “scrolls down” in the TABLE window
(set to display natural number X-values).
Hence x^2 == 150 (mod 1009)
*does* have a solution.
(In fact, two of them; the other is of course
870 (= 1009 – 139)).
In detail,
139^2 = 19321 = 19*1009 + 150
and
870^2 = 756900 = 750*1009 + 150.
(The point here is that one may easily
*verify* such results by “paper-&-pencil”
methods.)
To find the roots *entirely* by p-&-p methods,
the best line of attack would probably be via
“primitive roots”.