Download Theory of Numbers (V63.0248) Professor M. Hausner Answer sheet

Theory of Numbers (V63.0248)
Professor M. Hausner
Answer sheet for Homework 1. Sec. 1.2, p. 17 ff
1a, *1c (Show your work) , *3c (You can use the GCD spreadsheet here), 3d, 5, 6, 11, 12,
*14, 21, 25, 30, 31, 34, 35, 47D, 50D
Prove: If (a, b) = 1 and a|c and b|c, then ab|c.
1c. Find (2947,3997). Using the Euclidean algorithm, we get the following table of quotients
and remainders.
Numerator Denominator Quotient
3997
2947
1
2947
1050
2
1050
847
1
847
203
4
203
35
5
35
28
1
28
7
4
Remainder
1050
847
203
35
28
7
0
Thus (2947,3997)=7 using the Euclidean algorithm.
3c. Using the GCD spreadsheet calculation, 43x + 64y = 1 for x = 3, y = −2.
14. if n is odd, we have n = 2k + 1 for some integer k. So n2 = (2k + 1)2 = 4k 2 + 4k + 1 =
4k(k + 1) + 1. But the product of 2 consecutive numbers is even.1 So k(k + 1) = 2T , and
n2 = 8T + 1. This gives the result.
Sec. 1/3, p. 28 1, *2, 3, 5, 9, *11, 16, 18
2. A little experimentation shows that 4 consecutive numbers contains a number divisible
by 4. (In class, we showed in general that r consecutive numbers will contain a number
divisible by r.) On the other hand, we can find 3 consecutive square-free number, e.g 5 ,6
,7. So the largest size of a consecutive square-free sequence is 3. The same analysis Shows
that 8 consecutive numbers contains a number divisible by 8 – so will not be cube-free. The
numbers from 1 to 7 is a cube-free sequence, and so the answer here is 7.
11. If x and y are odd, then we have shown in problem 14 above, the x2 = 8S + 1 and
y 2 = 8T + 1. So x2 + y 2 = 8(S + T ) + 2. If x2 + y 2 = z 2 , z would have to be even. (If it
were odd, its square would be odd which is not possible since x2 + y 2 is even. So z = 2R
and z 2 = 4R2 . Equating x2 + y 2 and z 2 , we get 8(S + T ) + 2 = 4R2 . So 4(S + T ) + 1 = 2R2 .
This is impossible, because one side is even and the other is odd.
1
This is true if k is even. If it is odd, then k + 1 is even. So in any case k(k + 1) is even.