Hard math questions, again?

Question

1. Show that there are no positive integers n for which n^4+2n^3+2n^2+2n+1 is a perfect square. Are there any positive integers n for which n^4+2n^3+2n^2+2n+1 is a perfect square?

2. Can you find a positive integer n for which 1/2n is a perfect square, 1/3n is a perfect cube and 1/5n is a perfect fifth power?

3. In the game if Incan basketball, a points are given for a free throw and b points are given for a field goal, where a and b are positive integers. If a=2 and b=5, then it is not possible for a team to score exactly 1 or 3 points.

1)Are there any other unattainable scores?

2)How many unattainable scores are there if a=3 and b=5?

3)Is it true for any choice of a and b that there are only finitely many unattainable scores?

4)Suppose a and b are unknown, but it is known that neither a nor b is equal to 2 and that there are exactly 64 unattainable scores. Can you determine a and b?

Thanks in advance. I'm preparing for math team and these are old questions I didn't get.

Answer 1

1. Factor the expression n⁴ + 2n³ + 2n² + 2n + 1 =

= (n² + 1)² + 2n(n² + 1) = (n² + 1)² + 2n(n² + 1) + n² - n² =

= (n² + n + 1)² - n² = (n + 1)² (n² + 1)

Now since (n + 1)² is a perfect square, the above product to be a perfect square requires n² + 1 to be also a perfect square, the latter possible only if n=0, hence for n - positive integer the given expression can not be a perfect square.

The second sentence isn't clear - the same expression repeats.

2. If n/2 = x², n/3 = y³ and n/5 = z⁵ then

2x² = 3y³ = 5z⁵ and we should seek n as

n = 2^a * 3^b * 5^c, where a, b, c - naturals, such that:

a-1 ≡ b ≡ c ≡ 0 (mod 2);

a ≡ b-1 ≡ c ≡ 0 (mod 3);

a ≡ b ≡ c-1 ≡ 0 (mod 5), the smallest solution in natural numbers is

a = 15, b = 10, c = 6, so

n = 2¹⁵ 3¹⁰ 5⁶ = 30233088000000, indeed

n/2 = (2⁷ 3⁵ 5³)², n/3 = (2⁵ 3³ 5²)³, n/5 = (2³ 3² 5)⁵

3. Read my answer to a related question some time ago for explanation:

http://answers.yahoo.com/question/index;...

1) No, since (2 - 1)(5 - 1)/2 = 2.

2) Similarly (3 - 1)(5 - 1)/2 = 4, the 4 unattainable scores are 1, 2, 4, 7.

3) Yes, it's explained in the other answer - there are finitely many scores, less than ab.

4) 2 * 64 = 128 = (a - 1)(b - 1) = 4 * 32 = 8 * 16 are the only possibilities to remain since a > 2 and b > 2.