Indefinite Series, Perfect Squares
Date: 01/01/98 at 15:54:23 From: Connie Subject: Indefinite series/perfect squares. A table consists of eleven columns. Reading across the first row of the table, we find the numbers 1991, 1992, 1993, .... 2000, 2001. In the other rows, each entry in the table is 13 greater than the entry above it, and the table continues indefinitely. What is the number of columns containing at least one perfect square? I've tried approaching this question by finding the equation for the nth term for all of the columns, but that just adds to the confusion. Can you pleases help me solve this? Connie
Date: 01/08/98 at 10:01:41 From: Doctor Nick Subject: Re: Indefinite series/perfect squares. Hi Connie - interesting question! What it all boils down to is: if you have a perfect square, and divide it by 13, what can the remainder be? To answer this, use the fact that every integer n can be written as n = 13 a + b, where a and b are integers, and 0 <= b <= 12. That is, b is the remainder you get when you divide n by 13. Then, n squared is n^2 = (13 a + b)^2 = 169 a + 26 b + b^2. Now, if we divide n^2 by 13, we get 13 a + 2 b + (b^2)/13. Since 13 a + 2 b is a whole number, we can tell that the remainder we get when we divide n^2 by 13 is the same as the remainder we get when we divide b^2 by 13. Now we can make the following table: b remainder when b^2 is divided by 13 0 0 1 1 2 4 3 9 4 3 5 12 6 10 7 10 8 12 9 3 10 9 11 4 12 1 This tells us a lot: a perfect square must have a remainder of 0,1,3,4,9,10, or 12 when divided by 13. That means that if a number gives a remainder of 2,5,6,7,8, or 11 when divided by 13, then it's not a perfect square. Now, all of the numbers in the first column give a remainder of 2 when divided by 13: 1991 = 153*13+2, 2004 = 1991+13 = 154*13+2, etc. Therefore, none of the numbers in the first column can be perfect squares! The next column under 1992 is different. The remainder you get when you divide 1992 by 13 is 3, so this column might contain perfect squares. Looking at the table above, we can see that perfect squares that give a remainder of 3 when divided by 13 are squares of numbers that give a remainder of 3 or 9 when divided by 13. Now, the square root of 1992 is about 44; the next bigger number that gives 4 or 9 as its remainder is 48; 48 squared is 2304; and so 2304 is a perfect square in the second column (one way to recheck that it's in the second column is to check that 2304-1992 is a multiple of 13). Now continue for the rest of the columns: either eliminate them for having a remainder other than 0,1,3,4,9,10, or 12 when divided by 13, or find a square in the column. This is definitely a sophisticated question. Remember to have fun with it, and write back if you need more info. You could also look in a library for books on number theory. Look up "modular arithmetic" and "congruences" to find more on the method we used here. -Doctor Nick, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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