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1. Prove that if n is a positive integer such that the equation

      x3 - 3xy2 + y3 = n

has a solution in integers x, y, then it has at least three such solutions. Show that the equation has no solutions in integers for n = 2891.

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2. The function f(n) is defined on the positive integers and takes non-negative integer values. f(2) = 0, f(3) > 0, f(9999) = 3333 and for all m, n:

      f(m+n) - f(m) - f(n) = 0 or 1.

Determine f(1982).

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3. Find the values of m if:

 m2 + 9m = k2

 

& m & k are integers.

 

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4. Show that there are no integers a, b, c for which

 

a2 + b2 – 8c = 6

 

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5. Find 1.1! + 2.2! + ... + n.n!

 
 
 
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Solution

1. If x, y is a solution then so is y-x, -x. Hence also -y, x-y. If the first two are the same, then y = -x, and x = y-x = -2x, so x = y = 0, which is impossible, since n > 0. Similarly, if any other pair are the same.

2891 = 2 (mod 9) and there is no solution to x3 - 3xy2 + y3 = 2 (mod 9). The two cubes are each -1, 0 or 1, and the other term is 0, 3 or 6, so the only solution is to have the cubes congruent to 1 and -1 and the other term congruent to 0. But the other term cannot be congruent to 0, unless one of x, y is a multiple of 3, in which case its cube is congruent to 0, not 1 or   -1.

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2. We show that f(n) = [n/3] for n <= 9999, where [ ] denotes the integral part.

We show first that f(3) = 1. f(1) must be 0, otherwise f(2) - f(1) - f(1) would be negative. Hence f(3) = f(2) + f(1) + 0 or 1 = 0 or 1. But we are told f(3) > 0, so f(3) = 1. It follows by induction that f(3n) ≥ n. For f(3n+3) = f(3) + f(3n) + 0 or 1 = f(3n) + 1 or 2. Moreover if we ever get f(3n) > n, then the same argument shows that f(3m) > m for all m > n. But f(3.3333) = 3333, so f(3n) = n for all n <= 3333.

Now f(3n+1) = f(3n) + f(1) + 0 or 1 = n or n + 1. But 3n+1 = f(9n+3) ≥ f(6n+2) + f(3n+1) ≥ 3f(3n+1), so f(3n+1) < n+1. Hence f(3n+1) = n. Similarly, f(3n+2) = n. In particular f(1982) = 660.

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4.  a2 + b2 – 8c = 6

or, a2 + b2 = 6 + 8c

 

Right Hand Side is an even number.

 

Case I

If both a & b are even,

Let, a = 2m & b = 2n

 

4m2 + 4n2 = 8c + 6

2m2 + 2n2 = 4c + 3

2(m2 + n2) = 4c + 3

 

LHS is an even number, but RHS is an odd number.

Hence, both cannot be even.

 

Case II

If both a & b are odd,

Let, a = 2m+1 & b = 2n+1

(2m+1)2 + (2n+1)2 = 8c+6

m(m+1) + n(n+1) = 2c+1

 

The product of consecutive number is even

RHS is odd

 

Therefore both are not possible
 
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5. n.n! + (n-1).(n-1)! + (n-2)(n-2)! + ………3.3! + 2.2! + 1.1!

 

= (n+1-1).n! + (n-1).(n-1)! + (n-1-1).(n-2)! + ………(4-1).3! + (3-1).2! + (2-1).1!

 

= [(n+1).n! – n!] + [n.(n-1)! – (n-1)!] + [(n-1).(n-2)! – (n-2)!] + ………..[4.3! – 3!] + [3.2! – 2!] + [2.1! – 1!]

 

= {(n+1).n! + n.(n-1)! + (n-1).(n-2)! + …..4.3! + 3.2! + 2.1!} – {n! + (n-1)! + (n+2)! + …..3! + 2! + 1!}

 

= {(n+1)! + n! + (n-1)! + ……4! + 3!+ 2!} - {n! + (n-1)! + (n+2)! + …..3! + 2! + 1!}

 

= (n+1)! – 1!

 

= (n+1)! - 1

 

Solution by Kapil Goel


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