Difference between revisions of "2015 USAMO Problems/Problem 1"
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<math>2x - 3n^2 - 3n = \pm (2n^3 + 3n^2 -3n -2)</math> | <math>2x - 3n^2 - 3n = \pm (2n^3 + 3n^2 -3n -2)</math> | ||
− | <math>x = n^3 + 3n^2 - 1</math> or <math>x = n^3 + 3n | + | <math>x = n^3 + 3n^2 - 1</math> or <math>x = -n^3 + 3n + 1</math> |
− | Using substitution we get the solutions: <math>(n^3 + 3n^2 - 1, n^3 + 3n | + | Using substitution we get the solutions: <math>(n^3 + 3n^2 - 1, -n^3 + 3n + 1) \cup (-n^3 + 3n + 1, n^3 + 3n^2 - 1)</math> |
+ | |||
+ | ==Solution 2== | ||
+ | Let <math>n = \frac{x+y}{3}</math>. | ||
+ | Thus, <math>x+y = 3n</math>. | ||
+ | We have | ||
+ | <cmath>x^2+xy+y^2 = \left(\frac{x+y}{3}+1\right)^3 \implies (x+y)^2 - xy = \left(\frac{x+y}{3}+1\right)^3</cmath> | ||
+ | Substituting <math>n</math> for <math>\frac{x+y}{3}</math>, we have | ||
+ | <cmath>9n^2 - x(3n-x) = (n+1)^3</cmath> | ||
+ | Treating <math>x</math> as a variable and <math>n</math> as a constant, we have | ||
+ | <cmath>9n^2 - 3nx + x^2 = (n+1)^3,</cmath> | ||
+ | which turns into | ||
+ | <cmath>x^2 - 3nx + (9n^2 - (n+1)^3) = 0,</cmath> | ||
+ | a quadratic equation. | ||
+ | By the quadratic formula, | ||
+ | <cmath>x = \frac{1}{2} \left(3n \pm \sqrt{9n^2 - 4(9n^2 - (n+1)^3)} \right)</cmath> | ||
+ | which simplifies to | ||
+ | <cmath>x = \frac{1}{2} \left(3n \pm \sqrt{4(n+1)^3 - 27n^2} \right)</cmath> | ||
+ | Since we want <math>x</math> and <math>y</math> to be integers, we need <math>4(n+1)^3 - 27n^2</math> to be a perfect square. | ||
+ | We can factor the aforementioned equation to be <cmath>(n-2)^2 (4n+1) = k^2</cmath> | ||
+ | for an integer <math>k</math>. | ||
+ | Since <math>(n-2)^2</math> is always a perfect square, for <math>(n-2)^2 (4n+1)</math> to be a perfect square, <math>4n + 1</math> has to be a perfect square as well. | ||
+ | Since <math>4n + 1</math> is odd, the square root of the aforementioned equation must be odd as well. | ||
+ | Thus, we have <math>4n + 1 = a^2</math> for some odd <math>a</math>. | ||
+ | Thus, <cmath>n = \frac{a^2 - 1}{4}, </cmath> in which by difference of squares it is easy to see that all the possible values for <math>n</math> are just <math>n = p(p-1)</math>, where <math>p</math> is a positive integer. | ||
+ | Thus, <cmath>x+y = 3n = 3p(p-1).</cmath> | ||
+ | Thus, the general form for <cmath>x = \frac{1}{2} \left(3p(p-1) \pm \sqrt{4(p(p-1)+1)^3 - 27(p(p-1))^2} \right)</cmath> for a positive integer <math>p</math>. | ||
+ | (This is an integer since <math>4(p(p-1)+1)^3 - 27(p(p-1))^2</math> is an even perfect square (since <math>4(p(p-1)+1)</math> is always even, as well as <math>27(p(p-1))^2</math> being always even) as established, and <math>3p(p-1)</math> is always even as well. Thus, the whole numerator is even, which makes the quantity of that divided by <math>2</math> always an integer.) | ||
+ | Since <math>y = 3n - x</math>, the general form for <math>y</math> is just | ||
+ | <cmath>y = 3p(p-1) - \frac{1}{2} \left(3p(p-1) \pm \sqrt{4(p(p-1)+1)^3 - 27(p(p-1))^2} \right)</cmath> | ||
+ | (This is an integer since <math>4(p(p-1)+1)^3 - 27(p(p-1))^2</math> is an even perfect square (since <math>4(p(p-1)+1)</math> is always even, as well as <math>27(p(p-1))^2</math> being always even) as established, and <math>3p(p-1)</math> is always even as well. Thus, the whole numerator is even, which makes the quantity of that divided by <math>2</math> always an integer, which thus trivially makes <cmath>3p(p-1) - \frac{1}{2} \left(3p(p-1) \pm \sqrt{4(p(p-1)+1)^3 - 27(p(p-1))^2} \right)</cmath> an integer.) | ||
+ | for a positive integer <math>p</math>. | ||
+ | Thus, our general in integers <math>(x, y)</math> is <cmath>(\frac{1}{2} \left(3p(p-1) \pm \sqrt{4(p(p-1)+1)^3 - 27(p(p-1))^2} \right), 3p(p-1) - \frac{1}{2} \left(3p(p-1) \pm \sqrt{4(p(p-1)+1)^3 - 27(p(p-1))^2} \right).</cmath> | ||
+ | <math>\boxed{}</math> | ||
+ | |||
+ | -fidgetboss_4000 |
Latest revision as of 22:42, 15 March 2020
Problem
Solve in integers the equation
Solution
We first notice that both sides must be integers, so must be an integer.
We can therefore perform the substitution where
is an integer.
Then:
is therefore the square of an odd integer and can be replaced with
By substituting using we get:
or
Using substitution we get the solutions:
Solution 2
Let .
Thus,
.
We have
Substituting
for
, we have
Treating
as a variable and
as a constant, we have
which turns into
a quadratic equation.
By the quadratic formula,
which simplifies to
Since we want
and
to be integers, we need
to be a perfect square.
We can factor the aforementioned equation to be
for an integer
.
Since
is always a perfect square, for
to be a perfect square,
has to be a perfect square as well.
Since
is odd, the square root of the aforementioned equation must be odd as well.
Thus, we have
for some odd
.
Thus,
in which by difference of squares it is easy to see that all the possible values for
are just
, where
is a positive integer.
Thus,
Thus, the general form for
for a positive integer
.
(This is an integer since
is an even perfect square (since
is always even, as well as
being always even) as established, and
is always even as well. Thus, the whole numerator is even, which makes the quantity of that divided by
always an integer.)
Since
, the general form for
is just
(This is an integer since
is an even perfect square (since
is always even, as well as
being always even) as established, and
is always even as well. Thus, the whole numerator is even, which makes the quantity of that divided by
always an integer, which thus trivially makes
an integer.)
for a positive integer
.
Thus, our general in integers
is
-fidgetboss_4000