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Let's try this the old fashioned long division way..
sorry those are my random thoughts
don't have scrap paper on me
just gotta get the k in terms of n and 7.. hold on
im still thinkin btw haha
i think it's 1, just gotta prove it
okay, i think i got it: Suppose n=7a+k and n^2=7b+k
subtracting the two equations: n-n^2 = (7a+k) - (7b+k) or n-n^2 = 7a-7b = 7(a-b)
or n(n-1)=7(a-b). Now this means that either n is divisible by 7 or n-1 is divisible by 7.
Let a-b be some integer, so n(n-1) = 7i. As I've said either n is divisible by 7 or n-1 is divisible by 7 (whichever one's divisible by 7 would mean that the other one's divisible by i).
We know that n is not divisible by 7 (given), that means n-1 must be divisible by 7
Thus, if n-1 is some multiple of 7, n-1=7*x; or n = 7*x+1
which makes k = 1. Actually let's test this.
8/7 has remainder 1, so does 64/7
Try 30 and then 30^2 and test for equal remainders.
Use congruences k=1
n : 7 =x+k n^2 :7=y+k k=? n :7 =x+k --- n=x*7 +k ((x*7)+k)^2 :7=y+k 49x^2 +14kx +k^2 =7y+7k 14kx +k^2 -7k =7y-49x^2 k^2 +7k(2x-1) =7y-49x^2 k^2 +7(2x-1)k -7y+49x^2 =0 k^2 +7(2x-1)k -7(y-7x^2) =0 --- so now just you need to solve this quadratic equation for k_1 and k_2 .... - hope that is understandably !!! good luck bye