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1. Sep 23, 2008

### özgürden

from my turkish maths olympiads book

original question
http://www.akdeniz.edu.tr/fenedebiyat/math/olimpiyat/2006a.pdf [Broken]
question 19
^=exponent

we are looking for integer solutions (x, y)
x^3 - y^3 = 2.(y)^2 + 1
Find how many integer solutions there are to given equation that satisfy the given condition.

Last edited by a moderator: May 3, 2017
2. Sep 24, 2008

### özgürden

incidentally ,choise 5 :E) Sonsuz çoklukta
meaning: infinite

3. Sep 24, 2008

### CRGreathouse

The set of solutions is a superset of {(1, 0), (-2, -3)}, which shows that it's not D.

4. Sep 25, 2008

### uart

Yes CRG, and there's one more solution, making answer B the correct one.

Here's my rough solution.

Let x=(y+a) for some integer "a".

Then x^3 - y^3 = 3a y^2 + 3a^2 y + a^3

3a y^2 + 3a^2 y + a^3 = 2y^2 + 1 implies that,

(3a-2) y^2 + 3a^2 y + a^3-1 = 0 *

We want integer solutions, but clearly there can be no integer solutions if there are no real solutions. So investigate this first.

Reals solutions to * imply that 9a^4 >= 4(3a-2)(a^3-1), which re-arranges to

9a^4 >= 12a^4 - 8a^3 - 12a + 8

3a^4 <= 8a^3 + 12a - 8

Since "a" must be integer we can solve the above inequality numerically or by trial and error and find that a = 1 or 2 or 3 are the only possible values that can give rise to real solutions to *.

Investigate a=1.

y^2 + 3y + 0 = 0 has solutions y=0 and y=-3, giving two solutions (x,y) = (1,0) and (-2,-3)

Investigate a=2

4y^2 + 12y + 7 = 0

D = 12^2 - 4*4*7 = 32, is not perfect square so there are no rational (and hence no integer) solutions.

Investigate a=3

7y^2 + 27y + 26 = 0

D = 27^2 - 28*26 = 1, so there are rational and therefore perhaps integer solutions. Check.

y = (-27 +/- 1)/14, which gives one integer solution, y=-2 and hence (x,y) = (1,-2) is also part of the solution set.

Summary. There are 3 integer_pair solutions to the original equation. (x,y) = (-2,-3), (1,0) and (1,-2).

Last edited: Sep 25, 2008