### 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!)*

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”.