Uzasadnij, że reszta z dzielenia przez 3 sumy kwadratów trzech dowolnych kolejnych liczb naturalnych wynosi 2.
n, n+1, n + 2 - trzy kolejne dowolne liczby naturalne
n^2 + ( n +1)^2 + ( n +2)^2 = n^2 + (n^2 + 2n + 1) + (n^2 + 4n + 4 ) = 3 n^2 + 6n + 5
zatem
[ 3 n^2 + 6n + 5 ] / 3 = [ 3 n^2 + 6n + 3 + 2 ] / 3 = n^2 + 2n + 1 . r 2
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n, n+1, n + 2 - trzy kolejne dowolne liczby naturalne
n^2 + ( n +1)^2 + ( n +2)^2 = n^2 + (n^2 + 2n + 1) + (n^2 + 4n + 4 ) = 3 n^2 + 6n + 5
zatem
[ 3 n^2 + 6n + 5 ] / 3 = [ 3 n^2 + 6n + 3 + 2 ] / 3 = n^2 + 2n + 1 . r 2
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