Download Homework 2 (1) Prove that if n is an integer and n 2 is even, then n

Homework 2
(1) Prove that if n is an integer and n2 is even, then n is even.
Solution. Suppose n is odd but n2 is even. Since odd times odd is odd, we have a
contradiction.
(2) Show that b2 + b + 1 = a2 has no positive integer solutions.
Solution. Suppose that that a0 and b0 are positive integer solutions. Then the equality
implies that a0 > b0 ≥ 1. Now a little algebra yields
b0 + 1 = (a0 + b0 )(a0 − b0 ).
Since a0 > b0 ≥ 1, we have
a0 + b0 ≥ b0 + 2 and a0 − b0 ≥ 1.
This violates the equality that a0 and b0 satisfy.
(3) Start with the set {3, 4, 12}. In each step you may choose two of the numbers a, b
and replace them by 0.6a − 0.8b and 0.8a + 0.6b. Can you reach the goal (a) or (b) in
finitely many steps:
(a) {4, 6, 12}
√
(b) {x, y, z} with |x − 4|, |y − 6|, |z − 12| each less than 1/ 3?
Solution. Let {x, y, z} be the set at some point, the invariant is that x2 +y 2 +z 2 = 132 ,
i.e., the points lie on the sphere centered at the origin of radius 13. Why is this an
invariant? When in doubt, if you’re given points, think distance. Note 32 + 42 +
122 = 132 and suppose you have a, b, c at some point in the process. Then we know
a2 +b2 +c2 = 132 and if we replace as in the problem (0.6a−0.8b)2 +(0.8a+0.6b)2 +c2 =
a2 + b2 + c2 = 132 . So
(a) Since (4, 6, 12) is 14 units away from the origin, we’ll never get there.
(b) Note that the inequality implies that (x − 4)2 + (y − 6)2 + (z − 12)2 ≤ 1. In
other words, this point is less than one unit away from (4, 6, 12), a point which
is one unit away from the sphere of radius 13. So you’ll never reach such a point
(x, y, z).
(4) Start with the positive integers 1, . . . , 4n − 1. In one move you may replace any two
integers by their difference. Prove that an even integer will be left after 4n − 2 steps.
Solution. Claim: the number of odd integers in the list remains even at each step.
Suppose you take out two numbers. If they’re both even, then so is a − b and you’re
neither lost nor gained an odd number. If one is even and one is odd, you’ve lost an
odd one but there difference is odd so you’ve an odd one, too. If they’re both odd,
you’ve lost two odds and gained an even. In either case, the parity of the number of
odd numbers stays the same. Since we start with an even number of odd numbers
(there are 2n of them), we can never have one odd number left. Since we can have a
single number left, it has to be even.
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