1
$\begingroup$

Let $a_1=19$, $a_2=98$. For $n\ge1$, define $a_{n+2}$ to be the remainder of $a_n + a_{n+1}$ when it is divided by $100$. What is the remainder when $a_1^2+a_2^2+\cdots+a_{1996}^2$ is divided by 8? I got (working mod 8) that the terms of the sequence squared are congruent to ${1,4,1,1,0,1}$ in this specific order, I also proved that $a_{n+1}^2$ is congruent to $(a_n + a_{n-1})^2$ from this I proved that if two consecutive terms of the sequence squared are congruent to 1 and 4 respectively then the the 3rd term squared is congruent to 1 and the fourth term squared is congruent to 1 from this I have reached the conclusion that the fifth term squared term is congruent to 0 or 4 and I think to complete my proof I need to show that $a_{6n}$ is congruent to $a_{6n+1}$, how do I do this? This is from BMO 1 1998 question 2.

  • 0
    What do yo umean by "the remainder of $a_n$ and $a_{n+1}$ when it is divided by $100$"? Do you mean the remainder of the division $\frac{a_n+a_{n+1}}{100}$? – 2017-02-14
  • 0
    Yeah sorry my bad. – 2017-02-14
  • 0
    Is $a_1^2\ldots a_{1996}^2$ the product of all of those, or just a list? Because it it's the product, and if there are two or more even numbers in there, the result is guaranteed to be divisible by $8$. – 2017-02-14
  • 0
    No the sum of all of these. – 2017-02-14
  • 1
    The sum? You wrote it as a product. – 2017-02-14
  • 0
    It makes more sense to work with $a_n$ modulo $4$ (it's hard to control $a_n$ mod $8$ because the mod $100$ action can cause it to vary by $\pm 4$), and then show that the residue of $a_n^2$ mod $8$ depends only on the residue of $a_n$ mod $4$. – 2017-02-14
  • 0
    I feel so stupid for having not noticed that! Thanks so much – 2017-02-14

1 Answers 1

0

Clearly the sequence of a(n) follows the pattern (odd,even,odd) (odd,even,odd)... As 1998 is divisible by 3, the brackets repeat a whole number of times. 1998/3=666 which is divisible by 2 but not 4. As an odd number squared is odd, there are 4k (where k is odd) odd numbers in the sum.

As an even number squared is a multiple of 4, the sum of the even squares is a multiple of 2*4=8, say 8b. As 8b+4k= 4(mod 8) when k is odd, the answer is :

4