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Thank you so much for the encouraging words

Thank you so much. That is so encouraging.

It's typing error. The point is. ( 0, -3,,-3). Sorry guys

Sorry it is a typing error. The point is (0, -3, -3).

Thank you so much. Thats very encouraging.

Thank you so much

Thank you so much. Your kind words make me feel it was worth the effort to make the video.

Thank you so much and glad it helped you. Best wishes for your tests.

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2/17 is the legendre symbol. For 2/17 use the formula 2/p= (-1)^(p^2-1)/8 . p is prime. So 2/17 = (-1)^36 =1 For 6/23 first write 6/23= (2/23)(3/23) for 2/23 use above formula 2/23= (-1)^(23^2-1)/8 = (-1)^66 =1 For 3 /23 use 3/p =1 if p ≡ ±1(mod12) otherwise = -1. As 23 ≡ -1(mod12) 3/23 =1 Please check my video czcams.com/video/ExPjXxf63yk/video.html on Legendre symbol for all formulas

I do not think the theorem has a name . It is proved using induction.

(I) m1=3, m2=5 these are modulus. M=m1 × m2 = 3×5 . Now n1= M/m1 and n2= M/m2 . I hope its clear

As (9/13) = 3^2/13 we used (a^2)/p =1 where p is prime. For 2/17 use the formula 2/p= (-1)^p^2-1 .So 2/17 = (-1)^36 =1

Madam (-1)^17^2-1= ? Please tell this madam

2/p= (-1)^(p^2-1)/8 . p is prime.

2/p= (-1)^(p^2-1)/8 .

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Glad it helped! Thank you

You’re welcome 😊

Thank you

Thank you so much. Yes trying to.

Try Hensel's Lemma

Please check my video on solving diophantine euations in 2 variables. There I have explained how general solution is written in terms of a parameter t. Diophantine equations if solvable have infinite solutions and the different values of t give the infinite solutions

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Not all quadratic congruence have a solution. For a even no solution will exist. You can check by giving different values to x the original congruence is never satisfied. put x=2 , 2^2-2 is not divisible by 2. 3^2 -2 is not sivisible by 2. ... As x is the initial solution, it is found from x^2 = a(mod2). Take an example if we are solving a congruence x^2 = 7(mod2), you can see x=3 satisfies . When we put x=3 in x^2 =a+ b2, gives 3^2 = 7 +b2 ,implies b=1 as x=3 and b=1 , xy= -b(mod2) is 3y=-1(mod2) solve this congruence. Here we can see y=1 satisfies so y=1

yes thats correct should be 11. Typing error

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Glad it was helpful. Thank you.

Nevermind, I did it :D, make y=10-x and substitute in the first eq.

a|(bx +cy), therefore a|(bx +cy +fz +.....+nn) if it divides gcd. this is one of the first rule of divisibility, which is bezouts theorem

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Is this the way x^2 +y^2 = (r^2-s^2)^2 + (2rs)^2 = r^4 +s^4 -2r^2s^2 + 4r^2s^2 =r^4 +s^4 + 2r^2s^2 = (r^2+s^2)^2 =z^2

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