AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
. For each positive integer 
, we define its twist 
as follows: write as 
, where 
are non-negative integers and 
, then 
. For the positive integer 
, consider the infinite sequence 
where 
and 
is the twist of 
for each positive integer 
.
if and only if the remainder when is divided by 
is either or 
.
every next number (from the fifth number) is the unit number of the sum of the four numbers preceding it. Is there any cases that we get the numbers 
and 
in this series?
, the diagonals 
and 
meet at the point 
. We know that 
and 
. If we have 
prove that 
.![$P \in \mathbb Q[x,y]$](http://latex.artofproblemsolving.com/a/3/8/a384cbf38fd3a7f812b22baaeb0860e7f6eb26c0.png)
with 
and ![$\forall (a,b) \in \mathbb Z^2 \cap [0,d]^2, P(a,b) \in \mathbb Z$](http://latex.artofproblemsolving.com/a/6/5/a65ecb4bb2d9b5e6096f322c2de8bd11b945ed6a.png)
.
?
. Find .
be nonnegative real numbers such that 
. Prove that![\[
\frac{a}{4 - b} + \frac{b}{4 - c} + \frac{c}{4 - a} + \frac{1}{16}(1 - a)^2(1 - b)^2(1 - c)^2 \leq 1,
\]](http://latex.artofproblemsolving.com/5/b/d/5bd3349071e075519bd986c845c500125b7d46f8.png)
and determine all such triples 
where the equality holds.
such that 
is unbounded and![\[2f(m)f(n)-f(n-m)-1\]](http://latex.artofproblemsolving.com/0/f/7/0f77b5addbf8f5e603eb1b6923cdac795cd132e9.png)
is a perfect square for all integer 
. Then, we have that the equation becomes![\[(t+1)^3=(3t)^{2}-xy.\]](http://latex.artofproblemsolving.com/0/e/5/0e53a7cf6a0081583b512872b510b3e1f55555c7.png)
Now, we can make the substitution 
Thus, we have![\[(t+1)^3=9t^2-x(3t-x).\]](http://latex.artofproblemsolving.com/6/5/c/65ca1ca5700f2cbfdcda34e045ee383dabdf0019.png)
From this, we have:![\[t^3-6t^2+3t+1=x^2-3xt.\]](http://latex.artofproblemsolving.com/e/9/9/e99f9c838969ea9936791c5f860895f4d8aa5e45.png)
We can express this as a quadratic in 
:![\[x^2-3tx-(t^3-6t^2+3t+1).\]](http://latex.artofproblemsolving.com/7/5/a/75aec218a4a48992a3e92737c857e1f1dbf9c96b.png)
And now, by the quadratic formula, we have that the solutions to this equation are of the form![\[x=\frac{3t \pm \sqrt{9t^{2}+4t^3-24t^2+12t+4}}{2}\]](http://latex.artofproblemsolving.com/b/3/d/b3d00b489179f8ab47e9f9c2f5d9225bde774ea8.png)
![\[=\frac{3t \pm \sqrt{4t^3-15t^2+12t+4}}{2}\]](http://latex.artofproblemsolving.com/c/f/5/cf5753f0be4dacaefe34ed4f8e95649ed3c86315.png)
For this to be an integer, it's necessary for the expression in the square root to be a perfect square;
. is a square [which is what it factors to]
is a perfect square. However, this means that 
for some integral . Thus, we have that![\[(x,y)=\frac{3n^2+n\pm(n^2+n-2)(2n+1)}{2}\]](http://latex.artofproblemsolving.com/a/6/8/a68b19791caf8d0685cde51083d18857f59b7a48.png)
.![\[=(n^3+3n-1,-n^3+3n+1)\]](http://latex.artofproblemsolving.com/2/0/0/20054a708ecdbe05e83f9191816a4ee49750a138.png)
.
and 
for 
integral. Let 
which is an integer, then:
(since 
). This quadratic has an -discriminant:
so 
for some integer 
. Then is odd, so with the substitution 
we obtain 
. By the quadratic formula:
If 
, then 
and the first equation gives 
which are solutions included in the solution set..
, so 
. Since the equation is symmetric in 
, these give the general solutions as before.
and 
for all integers .
and 
. Then we obtain![\[\frac{3a^2+b^2}{4}=\left(\frac{a}{3}+1\right)^3\]](http://latex.artofproblemsolving.com/8/b/7/8b7a9d24e3d4157165490ab1a7077269f2141e45.png)
so we can set 
to get![\[\frac{3(9k^2-18k+9)+b^2}{4}=k^3\iff b^2=4k^3-27k^2+54k-27=(k-3)^2(4k-3).\]](http://latex.artofproblemsolving.com/1/b/9/1b95b9dc071f767d8cf2dfc482344f166c0930fe.png)
Then we must have 
for some integer , which means that 
and 
. Then 
, so that![\[(a,b)=(3n^2+3n,2n^3+3n^2-3n-2),(3n^2+3n,-2n^3-3n^2+3n+2)\]](http://latex.artofproblemsolving.com/c/6/d/c6d8dd137db93035ec29fc99bcbc882ad4b2d9a0.png)
and since 
and 
our final answer is![\[(x,y)=\boxed{(n^3+3n^2-1,-n^3+3n+1)},\boxed{(-n^3+3n+1,n^3+3n^2-1)}.\]](http://latex.artofproblemsolving.com/4/6/0/460dcb43ecc1f4c493b534d27063d941511a23c4.png)
, to get![\[27x^2+27xy+27y^2=(x+y+3)^3\]](http://latex.artofproblemsolving.com/b/4/e/b4eba7501d9b012f357859633b850c3f3c051771.png)
Expanding we get,![\[27x^2+27xy+27y^2 = (x+y)^3+9x^2+18xy+9y^2+27x+27y+27\]](http://latex.artofproblemsolving.com/4/c/3/4c3c753f86008569c56cf61914438f66c625588e.png)
We want another cubic term and a nicer LHS so we move the squares from the LHS to the RHS![\[27xy= (x+y)^3-18x^2+18xy-18y^2+27x+27y+27\]](http://latex.artofproblemsolving.com/b/d/4/bd4d2c94317222726f46caf19a029d2c26375bec.png)
Note that we can make 
because of the coefficients of the squares so we make it . Making the RHS, , we get![\[-27xy+81x+81y-243 = (x+y)^3-18x^2-36xy-18y^2+108x+108y+216 = (x+y-6)^3\]](http://latex.artofproblemsolving.com/f/f/2/ff29d32edc80f06150a64b6c90857a8fbbccf367.png)
The LHS can be factored by SFFT to get![\[-27(x-3)(y-3) = (x+y-6)^3\]](http://latex.artofproblemsolving.com/b/6/1/b617f27e34dd7e33e5ac8b8cf0ada0b8a8d8c7b0.png)
Substitute 
and 
to get![\[-27ab=(a+b)^3\]](http://latex.artofproblemsolving.com/1/3/f/13f7e8322c32ba0055c1dc5f69e4a41545d8847b.png)
Note that we have symmetric sums, so we let 
and 
and plug it into a quadratic to get![\[z^2-pz+q = z^2-pz-(\frac{p^3}{27}) = 0\]](http://latex.artofproblemsolving.com/d/a/0/da05dd3dcc9fe146af03aa8cf687c345189ced95.png)
If we use the quadratic formula, we get![\[(a,b) = \cfrac{p\pm\sqrt{p^2+\frac{4p^3}{27}}}{2}\]](http://latex.artofproblemsolving.com/0/f/d/0fdc757dc84bbe4f3be08ee92a5f295090dffbcf.png)
Note that 
must be divisible by 
, so we substitute 
to get![\[(a,b) = \cfrac{3k\pm k\sqrt{9+4k}}{2}\]](http://latex.artofproblemsolving.com/4/d/0/4d0199785582bcca2e5d682dbff8ca15de001c6e.png)
We need 
to be a square but it also has to be an odd square so we need![\[9+4k = (2n+1)^2\]](http://latex.artofproblemsolving.com/4/9/e/49e069bca8685006cbe611a4569c8fe41b68e35e.png)
Rearranging the terms give![\[k = n^2+n-2\]](http://latex.artofproblemsolving.com/c/4/c/c4ca470a0bb485443aa7eca4008cd56d6e3d83bd.png)
We can plug this back into the quadratic formula to get
We can also flip it because of symmetry. We still have 
so we add to each of the expressions to get
part.
substitution, or observing that a cubic polynomial just happens to have a squared factor. We'll present a motivated solution, using natural observations: ones that a contestant might first notice when approaching the problem.
factors: with this in mind, define 
and set
so![\[ x^2 + xy + y^2 = -ab. \]](http://latex.artofproblemsolving.com/6/d/d/6ddbec2d8ef5e0f01f3e010a7b81b73e02f9ac29.png)
In addition, we have
The provided equation thus transforms into![\begin{align*}
-ab & = \left( \frac{a+b}{3(\omega - \omega^2)} + 1 \right)^3\\
\iff -27ab(\omega - \omega^2)^3 & = [a + b + 3(\omega - \omega^2)]^3\\
\iff 27ab[3(\omega - \omega^2)] & = [a + b + 3(\omega - \omega^2)]^3
\end{align*}](http://latex.artofproblemsolving.com/4/f/7/4f7896879ae5acbd4545996ade411f700e603837.png)
where we observe that 
.
, so that![\[ (a+b+c)^3 = 27abc. \]](http://latex.artofproblemsolving.com/9/7/3/973bc3c3b4977a355dbca8e659e8e13de64f2265.png)
However, this equation can be easily solved:
if and only if there exist , 
, with sum 
satisfying 
.
factorization on steroids. 
, , by the same power of 
without affecting anything, so we can suppose 
. and satisfying
Moreover, these describe all complex solutions 
to the equation: take and with sum 
, and solve for and 
.
. Observe that![\[ \overline{q}^3 = \overline{\omega y - \omega^2 x} = \omega^2 y - \omega x = -p^3, \]](http://latex.artofproblemsolving.com/d/e/3/de3671412f153bbae0fa29e197346eece4075cd4.png)
so![\[ p^3 = -\overline{q}^3 \implies p = -\omega^k \overline{q} \]](http://latex.artofproblemsolving.com/3/1/e/31e5d6c4b39a5ca423f92d10c6d2731d92b1ddc9.png)
for some 
.
, then 
, so
a contradiction.
, we obtain a similar contradiction. 
; i.e. 
. In other words, and have opposite real parts and equal imaginary parts. From![\[ p + q = \omega - \omega^2 = 2\omega + 1, \]](http://latex.artofproblemsolving.com/b/8/a/b8ad4c82062c14fbc94fb50826d878471cb48d94.png)
we obtain that there exists a real satisfying
Finally, we can solve for and . We have
This is a linear system: the solution is![\[ (x, y) = (-t^3 + 3t + 1, t^3 + 3t^2 - 1). \]](http://latex.artofproblemsolving.com/e/e/2/ee2eb77b5d3cb01d7f5c76ea2efb5b02e49ec39f.png)
and are real. Thus, all real solutions are given by the mentioned![\[ (x, y) = (-t^3 + 3t + 1, t^3 + 3t^2 - 1) \]](http://latex.artofproblemsolving.com/3/0/8/308eede8b64feb24219459697e2a1989e2e790f8.png)
for real .
and 
are integers, then is also an integer. (The converse is clearly true.)![\[ x = -t^3 + 3t + 1 \quad \text{and} \quad y = t^3 + 3t^2 - 1 \]](http://latex.artofproblemsolving.com/5/9/f/59f416153a6c31b2927ee34500eb9308de11187c.png)
as ranges over the integers. Their validity can be seen by reversing the steps above.
, so set 
and 
. Then, the given equation becomes![\[ 9s^2 - p = (s + 1)^3 = s^3 + 3s^2 + 3s + 1\implies p = -s^3 + 6s^2 - 3s - 1.\]](http://latex.artofproblemsolving.com/d/8/3/d83116ca89b0d2e0bcb54007eb889803ff51ac11.png)
Now, notice that are the roots of 
. Since are integers, the discriminant of must be a perfect square. The discriminant is![\[ D = 9s^2 - 4(-s^3 + 6s^2 - 3s - 1) = 4s^3 - 15s^2 + 12s + 4 = (s - 2)^2(4s + 1),\]](http://latex.artofproblemsolving.com/3/e/a/3ead6cc32345828bff9f8fa2c83be6c50378c0ce.png)
which is a square if and only if 
is a square. Thus, set 
where is an integer. Then, we have 
So the roots of are![\[ \frac{3n^2 + 3n\pm (2n^3 + 3n^2 - 3n - 2)}{2} = n^3 + 3n^2 - 1, -n^3 + 3n + 1.\]](http://latex.artofproblemsolving.com/b/7/8/b786a9d8a60ba90b06da5e228968c88ff097f970.png)
Hence, the answer is 
(and permutations) for any integer .
, with equality when x=3, y=3.
, equality when 
. The solutions are 
.
and substitute these in to get
Taking 
of this gives 
, so we let 
. into the equation, we get that
Giving that 
is a (odd) perfect square, so we let it be 
![\[(x,y)=\left(\frac{3\left(\frac{m^2-1}{4}\right)+m\left(\frac{m^2-1}{4}\right)}{2},\frac{3\left(\frac{m^2-1}{4}\right)-m\left(\frac{m^2-1}{4}\right)}{2}\right),\]](http://latex.artofproblemsolving.com/f/5/b/f5bc6bd6c5aed579bba255156ebb6611ec51c473.png)
or![\[(x,y)=\left(\frac{3\left(\frac{m^2-1}{4}\right)-m\left(\frac{m^2-1}{4}\right)}{2},\frac{3\left(\frac{m^2-1}{4}\right)+m\left(\frac{m^2-1}{4}\right)}{2}\right),\]](http://latex.artofproblemsolving.com/d/9/7/d973ccee027890383d2b5c99339d4bc8067832b4.png)
where 
is an odd integer.
. Let 
. Now, we have that![\[(x+y)^2-xy=x^2+xy+y^2=\left(\frac{x+y}{3}\right)^3,\]](http://latex.artofproblemsolving.com/9/3/7/937c77c3ddc9273b58a220a65d91b2f6990e8146.png)
![\[\iff 9k^2-xy=(k+1)^3,\]](http://latex.artofproblemsolving.com/c/6/f/c6f2d442811d396d83771fd8f838c2afb7c34049.png)
![\[\iff xy=9k^2-(k+1)^3,\]](http://latex.artofproblemsolving.com/5/b/c/5bceaa260759bdc7eb99ec26f783d6d3d33562d9.png)
and if we put this into a quadratic equation, note that , must be roots of![\[r^2-(3k)r+(9k^2-(k+1)^3),\]](http://latex.artofproblemsolving.com/b/1/8/b1819caef5964afffb3a3dc7a4568a2a0dd97c9d.png)
which has discriminant 
. Therefore, if , are integers, we must have 
be a perfect square. However, this factors as 
, which means that must be a perfect square! Therefore, if we let 
(note that must be odd for to be an integer; and must be an integer since and !) and plug it in, we get![\[r=\frac{3\left(\frac{m^2-1}{4}\right)\pm m\left(\frac{m^2-1}{4}\right)}{2},\]](http://latex.artofproblemsolving.com/3/8/c/38c5f2cc4235df9fc6cc8cda7b9f2c9c03d2070a.png)
which must be an integer since all odd squares are mod 
, making 
even. Therefore our solutions are all in the formoras desired, finishing the problem.
.
is odd, the LHS is odd but the RHS is even, a contradiction. Hence is is even. Clearly . Let 
and 
, which are integers. Then![\[
27a^2+b^2=(2a+1)^3=8a^3+12a^2+6a+1\implies b^2=(a-1)^2(8a+1).
\]](http://latex.artofproblemsolving.com/2/d/e/2dec44ccac9b722d39ca0fb75bd9beff65f10c54.png)
It follows that 
is a perfect square so 
. Since 
and 
have the same parity, there exists 
such that 
and 
. Thus 
so 
. It follows that 
so![\[
(x,y)=\left(\frac{3d(d+1)}{2}+\frac{(d-1)(d+2)(2d+1)}{2},\frac{3d(d+1)}{2}-\frac{(d-1)(d+2)(2d+1)}{2}\right)=\boxed{(d^3+3d^2-1,-d^3+3d+1)}
\]](http://latex.artofproblemsolving.com/d/c/a/dcac51ab7f02cd07d9689171ff755713e49a6cb3.png)
are the only solutions. 
![\[(x, y) = (n^3 + 3n^2 - 1, -n^3 + 3n + 1)\]](http://latex.artofproblemsolving.com/5/7/0/570461520b8b531d813606f96b85d7a920e2372b.png)
for any 
. It is not hard to check that this works. We will now show that a pair must satisfy this form. Since the LHS is an integer, this is enough to imply that 
. Set 
, 
for integers 
. Substituting, we obtain![\[p = 9j^2 - (j + 1)^3\]](http://latex.artofproblemsolving.com/e/c/5/ec5dc36d07e8f020a8a8f87b87cd623e84243c6c.png)
However, 
are the roots of 
, so the discriminant![\[\delta = 4(j + 1)^3 - 27j^2 = (4j + 1)(j-2)^2\]](http://latex.artofproblemsolving.com/2/5/b/25b5dcb8580db048c737a4f719f05dbb5240213d.png)
is a perfect square, which is equivalent to 
. By the quadratic formula, the solutions to are![\[\frac{3(n^2 + n) \pm (n^2 + n - 2)(2n + 1)}{2} = \boxed{(n^3 + 3n^2 - 1, -n^3 +3n + 1)}\]](http://latex.artofproblemsolving.com/0/8/0/0804b6b735f7db6820c81491842ddf8d73614cdf.png)
and we are done.
and by parity inspection both are odd. Now consider the substitution 
for integers , the relation becomes![\[(3a - b)^2 + (3a - b)(3a + b) + (3a + b)^2 = (2a + 1)^3 \iff 8a^3 - 15a^2 + 6a + 1 = b^2\]](http://latex.artofproblemsolving.com/d/1/1/d11dbab5e69b80912f878510a4cca88ca9dcba89.png)
which factors as 
. Now 
must be a perfect square, which can all be determined by 
for some odd . Letting 
yields![\[ (a, b) = \left( \dfrac{k^2 - 1}{8}, \dfrac{k^3 - 9k}{8} \right) \implies (x, y) = \left( \dfrac{-k^3 + 3k^2 + 9k - 3}{8}, \dfrac{k^3 + 3k^2 - 9k - 3}{8} \right) \]](http://latex.artofproblemsolving.com/6/8/4/6846ca9131b92932d36596003e1d1f9c4f12f8c9.png)
as the entire curve of solutions for odd .
as 
![\[
x^2-3kx-k^3+6k^2-3k-1=0
\]](http://latex.artofproblemsolving.com/1/c/1/1c17426d0b97dc3fe69b4eb5903f3b6e83f3ca41.png)
Hence, 
, forcing that must be an odd perfect square.
![\[
\begin{cases}
x=t^3+3t^2-1\\
y=-t^3+3t+1\\
\end{cases}
\]](http://latex.artofproblemsolving.com/9/a/8/9a835a90336c902eaa1bcfcc4c8a095bc4fd5356.png)
or![\[
\begin{cases}
x=-t^3+3t+1\\
y=t^3+3t^2-1\\
\end{cases}
\]](http://latex.artofproblemsolving.com/4/e/c/4ecf8b925f4454aef7fb2946a30314eebf194279.png)
Here, 
![\[ x^2+xy+y^2 = \left(\frac{x+y}{3}+1\right)^3. \]](http://latex.artofproblemsolving.com/5/6/b/56b8cb8c13616501dd8308989cea2b338e91598b.png)
.
and 
.![\[ \frac14\left( (a+b)^2+(a+b)(a-b)+(a-b)^2 \right) = \left( \frac a3 + 1 \right)^3 \]](http://latex.artofproblemsolving.com/9/1/a/91a0ffa3f678c8b8a3ec7b1fe6c9f9098f428849.png)
where 
have the same parity. This becomes ![\[ 3a^2+b^2 = 4\left( \frac a3 + 1 \right)^3 \]](http://latex.artofproblemsolving.com/8/f/3/8f37b905da01aa58e7c519e8bdaf11525468db48.png)
which is enough to imply 
,
.![\[ b^2 = (c-2)^2 (4c+1). \]](http://latex.artofproblemsolving.com/9/5/c/95c544b526acc91915b18e99a24957ea27705078.png)
So a solution must have 
,![\[ x = \frac 18 \left( 3 (m^2-1) \pm (m^3-9m) \right) \quad\text{and}\quad y = \frac 18 \left( 3 (m^2-1) \mp (m^3-9m) \right). \]](http://latex.artofproblemsolving.com/5/3/9/539f1df030aef352f7350aaf4fdd945bb39637aa.png)
For mod ![\[
\boxed{x = n^3+3n^2-1
\quad\text{and}\quad
y = -n^3+3n+1}
\]](http://latex.artofproblemsolving.com/a/1/2/a122147172996eaad99bba75933f0d43a896e8be.png)
and permutations.
this gives the first few solutions of 
,
,
,
,
?
I thought this was NUMBER THEORY so mods, primes, even Eisenstein integers...
,
,
.![\[ (p^3 + q^3 + r^3)^3 = (3pqr)^3 \]](http://latex.artofproblemsolving.com/e/7/c/e7c8b86abbd851132c0f7a42d47a298d47ef0c22.png)
and so![\[ p^3 + q^3 + r^3 = 3pqr \cdot \omega^k \]](http://latex.artofproblemsolving.com/c/8/5/c85d296cc0ec078a9d8c345e6b8572df1aaf0576.png)
for a suitable 
,![\[ (p+q+r)(p + q\omega + r\omega^2)(p + q\omega^2 + r\omega) = 0. \]](http://latex.artofproblemsolving.com/8/a/e/8ae5a2a9ff3506e9f5ff8f57194c506e370b067d.png)
The existence of suitable 
,
a contradiction.
Similarly, we can find 
for convenience. In particular, 
and 
are nonzero integers, say 
is an integer 
.
From above, we also have![\[ 0 = 3t^2 + 3t + (m - n). \]](http://latex.artofproblemsolving.com/d/e/9/de98af1f73866f22d9bc681fe73551c97cf19d42.png)
We now consider two cases. Suppose 
.
as 
.
.
,
term, learning that 
.
.
.
.
,
.
We now apply the quadratic formula to obtain 
factors nicely into 
must be a perfect odd square.
.
.
.
to get 
(the negative solution just switches 
so 
.
.
and 
.