I managed to show $x-y$ is even, then rip

Let $k=\gcd(x,y)$, and let $x=ka, y=kb$ (note $\gcd(a,b)=1$). Then $k^6(3a^3+ab^2)(3b^3+ba^2)=k^7(a-b)^7$ so \[(3a^3+ab^2)(3b^3+ba^2)=k(a-b)^7.\] We may rewrite this as \[ab(3(a-b)^2+2b(a-b)+4ab)(3(a-b)^2-2a(a-b)+4ab)=k(a-b)^7.\] Now $\gcd(a-b,ab)=1$ so $a-b$ must divide $4ab$; else there is a prime that has higher exponent in the factorization of $a-b$ than in $4ab$, and this prime will divide the right hand side more than the left. But then $a-b$ must be $4,2$ or $1$. If $a-b$ is 2 or 4, then both are odd, and we can check using mod 8 that 16 is the highest power of two that can divide the left hand side, contradiction. Thus $a=b+1$ and using $ab(3a^2+b^2)(3b^2+a^2)=k(a-b)^7$ we get \[k=b(b+1)(3(b+1)^2+(b^2))(3b^2+(b+1)^2)\] so \[(x,y)=(ka, kb)=(b(b+1)^2(3(b+1)^2+(b^2))(3b^2+(b+1)^2), b^2(b+1)(3(b+1)^2+(b^2))(3b^2+(b+1)^2))\]is the whole curve of solutions (we can quickly check that this works for positive $b$) so we are done. (The smallest solutions, at $b=1,2$, are $(364,182)$ and $(11718, 7812)$.)

Personally, I set $p = x-y$ and $q = xy$, and after a three-page long bash, my solution set was \[(x, y) = \left( n(n+1)^2(16n^4+32n^3+28n^2+12n+3), n^2(n+1)(16n^4+32n^3+28n^2+12n+3) \right).\]This was over all positive integers $n$.

$50 and an icecream cake that Titu wrote this

Let $x=da$, $y=db$, where $\gcd(a,b) = 1$ and $a > b$. The equation becomes \[ (a-b)^7 \mid ab\left( a^2+3b^2 \right)\left( 3a^2+b^2 \right) \]with the ratio of the two becoming $d$. Note that if $a$ and $b$ are both odd, then $2^4$ exactly divides right-hand side contradiction. But now \[ \gcd(a-b,a) = \gcd(a-b,b) = \gcd(a-b, a^2+3b^2) = \gcd(a-b, 3a^2+b^2) = 1. \]So we conclude $a-b = 1$ which indeed works. These describe all solutions (up to permuting $x$ and $y$). Remark: For cosmetic reasons, one can reconstruct the curve explicitly by selecting $b = \frac{1}{2} (n-1)$, $a = \frac{1}{2} (n+1)$ with $n > 1$ an odd integer. Then $d = ab(a^2+3b^2)(3a^2+b^2) = \frac{(n-1)(n+1)(n^2+n+1)(n^2-n+1)}{4} = \frac{n^6-1}{4}$, and hence the solution is \[ (x,y) = (da, db) = \left( \frac{(n+1)(n^6-1)}{8}, \frac{(n-1)(n^6-1)}{8} \right). \]The smallest solutions are $(364,182)$, $(11718, 7812)$, \dots.

Guys, the left hand side is [(x+y)^6-(x-y)^6]/4--->trivial

This problem is essentially the same as 2015 J2 oops. Let $a=x+y,b=x-y$. Our equation can be rewritten as $$\frac{a^2-b^2}{4}\left(\left(\frac{a+b}{2}\right)^2+3\left(\frac{a-b}{2}\right)^2\right)\left(3\left(\frac{a+b}{2}\right)^2+\left(\frac{a-b}{2}\right)^2\right)=b^7\iff$$$$(a-b)(a+b)(4a^2+4b^2-4ab)(4a^2+4b^2+4ab)=64b^7\iff$$$$(a^3-b^3)(a^3+b^3)=4b^7\iff$$$$a^6=b^6(4b+1)$$. We conclude that $4b+1=(2n+1)^6$ for some integer $n$. Solving for $b$, $$b=\frac{1}{4}\times(4n^2+4n+1-1)((2n+1)^4+(2n+1)^2+1)=$$$$n(n+1)((4n^2+4n+1+1)^2-(2n+1)^2)=$$$$n(n+1)(4n^3+6n+3)(4n^2+2n+1)$$. We find $a=(2n+1)b$. It follows that $x=\frac{a+b}{2}=\boxed{n(n+1)^2(4n^3+6n+3)(4n^2+2n+1)}$ and $y=\boxed{n^2(n+1)(4n^3+6n+3)(4n^2+2n+1)}$. remark: this is philosophically identical to 2015 j2, same trick, slightly different finish, etc. i concur that this is t2 problem (probably a good idea to hunt down mathematical reflections problems now?)

Is there anybody who solved a) but not b)?

Wait yeah I considered taking gcd at first but got nowhere, then did the substitution mentioned in post 8 and it became super trivial might have screwed up part b though

wait guys ok use droid's substitution and get $x-y|16$ by taking mod $x-y.$ If $c=x-y,$ I got $a(a+c)(4a^2+6ac+c^2)(4a^2+2ac+c^2)=c^7d$ for some integer $d.$ For $c=2$ case, I used the fact that $a$ must be even and let $a=2a'$ to get the $c=1$ case. Is it ok if I didnt mention that there are no solutions, and if I just said that every solution to this is a subset of the solutions in the $c=1$ case?

I found the solution set but didn't prove that they were the only possible solutions. How many points do i get?

Whew! At least I got this one correct. Grrrr why did I not trig bash for #3.

probably just 2 pts for part a imo

Wait but this problem is trivial if you realize the left hand side is [(x+y)^6-(x-y)^6]/4, cuz 4(x-y)+1 must be a sixth power.

skipiano wrote: Is there anybody who solved a) but not b)? yea I don't get this either, it seems that solving (a) makes (b) obvious...

I got $x=\frac{((1+2k)^6-1)(k+1)}{4}, y=\frac{((1+2k)^6-1)(k)}{4}$

There might have been another way to generate solutions other than the substitution? I feel complex numbers could be used. Also I got the above as well.

Will I get credit for saying that $(x,y)=(\frac{b\sqrt[6]{4b+1}+b}{2},\frac{b\sqrt[6]{4b+1}-b}{2})$ for all $b$ such that $4b+1$ is a perfect sixth power? I didn't put it in terms of $n$

wait rip in an attempt to lose fewer points, I wrote that the smallest solution was $(182, 91)$ but an above post says it's $(364, 182)$? will I lose points? @below I don't think we have to but doing it won't hurt, unless if ur off by a factor of 2 if I get so lucky that I miss mop by 1 point because of this I'll rage so much

AopsUser101 wrote:

Also, I have a question: the tag in the problem contained “linear algebra”--can anyone please write up a linear algebra solution? you need to prove that your $x$ and $y$ are integers as well

Thanks @franchester. Fixed now!

We will prove both a and b at the same time. We will substitute $a=x+y,b=x-y$. However, to do this, we first rewrite the RHS of the equation: \begin{align*} (3x^3+xy^2)(x^2y+3y^3)&=xy(3x^2+y^2)(x^2+3y^2)\\ &=xy(3x^4+10x^2y^2+3y^4) \end{align*}With the ultimate goal of substituting $a,b$, we are tempted to rewrite $3x^4+10x^2y^2+3y^4$ in the form of $m(x^2+y^2)^2+n(x^2-y^2)^2$. A bit of algebra gives us $m=4,n=-1$, so we get: \begin{align*} xy(3x^4+10x^2y^2+3y^4)&=xy(4(x^2+y^2)^2-(x^2-y^2)^2) \end{align*}Now we write $x^2+y^2$ in the form $m(x+y)^2+n(x-y)^2$ so we can put in $a,b$. Simple algebra gets $x^2+y^2=\frac{(x+y)^2+(x-y)^2}{2}$, so we have: \begin{align*} xy(4(x^2+y^2)^2-(x^2-y^2)^2)&=xy(((x+y)^2+(x-y)^2)^2-(x+y)^2(x-y)^2) \end{align*}We note that $x=\frac{a+b}{2},y=\frac{a-b}{2}$, so we can rewrite the equation as: \begin{align*} (a^2-b^2)((a^2+b^2)^2-a^2b^2)&=4b^7\\ (a^2-b^2)(a^4+a^2b^2+b^4)&=4b^7\\ a^6-b^6&=4b^7 \end{align*}Which is quite nice. Take $\pmod{b^6}$ on both sides to get $a^6 \equiv 0 \pmod{b^6}$, so $b^6 \mid a^6 \implies b\mid a$. Thus let $a=bk$ where $k>2$ is a positive integer ($k=1$ gives $y=0$). The case where $x<y$ is dealt with by symmetry. The equation then rewrites: $$k^6=4b+1$$It becomes clear that $k^6$ must be odd, and any such $k$ would work. We can substitute $k=2n+1$ for a positive integer $n$. We get: $$b=\frac{(2n+1)^6-1^6}{4}=\frac{((2n+1)^3-1)((2n+1)^3+1)}{4}=\frac{2n(2n+2)(4n^2+6n+3)(4n^2+2n+1)}{4}=n(n+1)(4n^2+6n+3)(4n^2+2n+1)$$so $a=n(n+1)(2n+1)(4n^2+6n+3)(4n^2+2n+1)$ Using $x=\frac{a+b}{2},y=\frac{a-b}{2}$, we get the solution set as: $$\boxed{(x,y)=\left(n(n+1)^2(4n^2+6n+3)(4n^2+2n+1),n^2(n+1)(4n^2+6n+3)(4n^2+2n+1)\right)}$$and permutations, where $n$ is a positive integer. Since all steps were reversible (they were just manipulations and substitutions) we conclude that these describe all solutions. $\blacksquare$

Wait how did @InCtrl get from InCtrl wrote: $$a^6=b^6(4b+1)$$ to InCtrl wrote: $$4b+1=(2n+1)^6$$for some integer $n$. btw, here is a direct link to @InCtrl's post: https://artofproblemsolving.com/community/c5h1433989p8108553

we have $a=b(4b+1)^{1/6}$ so if a is integer then 4b+1 is perfect odd 6th power so the conclusion is valid regards leonard

We will prove part $(b)$, as it obviously implies part $(a)$. As in 2015 jmo #2, we set $a=x+y$ and $b=x-y$. The equation can be rewritten as \[xy(3x^2+y^2)(3y^2+x^2)=\frac{a^2-b^2}{4}(\frac{3(a+b)^2}{4}+\frac{(a-b)^2}{4})(\frac{3(a-b)^2}{4}+\frac{(a+b)^2}{4})=b^7\]. So \[(a-b)(a+b)(3(a+b)^2+(a-b)^2)(3(a-b)^2+(a+b)^2)= (a-b)(a+b)(4a^2+4b^2+4ab)(4a^2+4b^2-4ab)=16(a-b)(a+b)(a^2+b^2+ab)(a^2+b^2-ab)=16(a^3-b^3)(a^3+b^3)=64b^7\implies (a^3-b^3)(a^3+b^3)=a^6-b^6=4b^7\implies a^6=b^6(4b+1)\] Thus, we know that $4b+1=m^6$ for some odd integer $m$. We get $b=\frac{m^6-1}{4}$ and $a=\pm bm=\pm \frac{m^7-m}{4}$. Also note that $b+a=2x$ and $b-a=-2y$. So $2x=\frac{1}{4}((m^6-1)\ \pm\ m^7-m)$ and $-2y=\frac{1}{4}((m^6-1)\ \mp\ m^7-m)$ for odd integers $m$. Setting $m=2n+1$ here gives \[2x=(16n^6+48n^5+60n^4+40n^3+15n^2+3n)\ \pm\ (32n^7+112n^6+168n^5+140n^4+70n^3+21n^2+3n)\]and \[-2y=(16n^6+48n^5+60n^4+40n^3+15n^2+3n)\ \mp\ (32n^7+112n^6+168n^5+140n^4+70n^3+21n^2+3n)\] Thus, the answer is \[\boxed{(x,y)=(16n^7+64n^6+108n^5+100n^4+55n^3+18n^2+3n, 16n^7+48n^6+60n^5+40n^4+15n^3+3n^2)}\]for positive integers $n$.

Some details not included, specifically at $*$. Clearly, $(b)$ implies $(a),$ so following proves $(b)$. We will substitute $a=x+y$, $b=x-y$. Note that this means that $x=\tfrac{a+b}{2}$ and $y=\tfrac{a-b}{2}$. [call this identity 1.] Then, we have \[(3x^3+xy^2)(x^2y+3y^3)=b^7.\]\[=3x^5y+10x^3y^3+3xy^5.\]\[=xy(3x^4+10x^2y^2+3y^4).\]\[=(\frac{a^2-b^2}{4})(\frac{3(a+b)^2}{4}+\frac{(a-b)^2}{4})(\frac{3(a-b)^2}{4}+\frac{(a+b)^2}{4})=b^7.\]$*$ After doing a lot of expanding out, we get the relation \[16(a^3-b^3)(a^3+b^3)=64b^7.\]which means that \[a^6-b^6=4b^7\]or, \[a^6=b^6(4b+1).\]This means that $4b+1$ must be a perfect 6th power. Setting $4b+1=k^6$, we get that $b=\frac{k^6-1}{4}.$ Thus, we also have that $a=\pm \frac{k(k^6-1)}{4}.$ We then have, by identity 1, that \[x=\frac{1}{8}((k^6-1)\pm k(k^6-1)), y=\frac{-1}{8}((k^6-1) \mp (k(k^6-1))\]for odd $k$. Now, since $k$ is odd, let $k=2n+1$. Then, we are able to derive the final solutions for positive integers $n;$ these are \[(x,y)=(16n^7+64n^6+108n^5+100n^4+55n^3+18n^2+3n, 16n^7+48n^6+60n^5+40n^4+15n^3+3n^2).\]

djmathman wrote: 50 dollars and an icecream cake that Titu wrote this Relatable even more than the real life memes USAMO 2015 #1 says that assume $x=\frac{a+b}{2},y=\frac{a-b}{2}$ something will happen Plugging this into the equation , $xy(3x^2+y^2)(3y^2+x)=(x-y)^7$ we get \begin{align*} b^7&=\frac{1}{4}(a+b)(a-b)\frac{1}{4}(3(a+b)^2+(a-b)^2)\frac{1}{4}(3(a-b)^2+(a+b)^2)\\ &=\frac{1}{4}(a+b)(a-b)\frac{1}{4}(4a^2+4b^2+4ab)\frac{1}{4}(4a^2+4b^2-4ab)\\ &=\frac{1}{4}(a+b)(a^2+b^2+ab)(a-b)(a^2+b^2-ab)=\frac{1}{4}(a^3-b^3)(a^3+b^3)\\ &=\frac{1}{4}(a^6-b^6) \implies 4b^7+b^6=a^6 \implies b^6(4b+1)=a^6 \\ \implies 4b+1&=(2k+1)^6=64k^6+192k^5+240k^4+160k^3+60k^2+12k+1 \text{ for some } k>0\\ \implies b &=16k^6+48k^5+60k^4+ 40k^3+15k^2+3k \end{align*}Also , $$a=b(2k+1) \implies a=32k^7+112k^6+168k^5+140k^4+70k^3+21k^2+3k$$Thus the format of the solutions become, $$(x,y)=(\frac{a+b}{2},\frac{a-b}{2})=(16k^7+64k^6+108k^5+100k^4+55k^3+18k^2+3k,16k^7+48k^6+60k^5+40k^4+15k^3+3k^2)$$Note that $k \in \mathbb{N}$ since $k=0$ totally fails. One of the solutions are $(x,y)=(364,182)$(Too lazy to find out what we get for $k=2$ )

Let $a=x+y$ and $b=x-y$, then the equation miraculously turns into: $$a^6=4b^7+b^6.$$Now $4b+1=k^6$ for some $k$. Noticing $k$ is odd, we can let $k=2t+1$ and expand to get: $$b=\frac{(2t+1)^6-1}4\text{ and }a=kb=\frac{(2t+1)^7-2t-1}4.$$Solving for $x$ and $y$, we have: $$x=\frac{(2t+1)^7+(2t+1)^6-2t-2}8\text{ and }y=\frac{(2t+1)^7-(2t+1)^6-2t}8$$which we can easily (*) verify (by expansion) are integers and satisfy the original equation.

jasper be on that grind

lol this problem is so weird Notice that \[ (3x^3 + xy^2)(x^2y +3y^3) = 10x^3y^3 + 3xy^5 + 3x^5y = \frac{1}{4}\left((x + y)^6 - (x-y)^6\right).\]Thus, let $n =x + y$ and $m = x - y$, so we have \[ \frac{1}{4}(n^6 - m^6) = m^7\implies 4m^7 + m^6 = n^6.\]It follows that $4m + 1$ is a perfect sixth power, which we will call $k^6$. However, notice that $k^6 =4m + 1\equiv 1\pmod{4}$, so $k$ is odd. Hence, $m = \tfrac{k^6 - 1}{4}$ and consequently $n = km = \tfrac{k^7 - k}{4}$. Hence, \[ (x,y) = \left(\frac{m + n}{2}, \frac{n-m}{2}\right) = \left(\frac{k^7 + k^6 - k - 1}{8}, \frac{k^7 - k^6 - k + 1}{8}\right)\]where $k$ is any odd integer. This implies that there are infinitely many solutions, and we have shown that all solutions are of the form described above, so we are done.

Let $d = \gcd(x,y)$ such that $x = da$ and $y =db$ for $\gcd(a,b) = 1$. Substituting this in and simplifying yields $$ab(3a^2+b^2)(a^2+3b^2) = d(a-b)^7$$which implies that $$a-b \mid ab(3a^2+b^2)(a^2+3b^2)$$However, since $\gcd(a,b) = 1$ we know that $\gcd(a-b,a) = \gcd(a-b,b) = 1$ so actually $$a-b \mid (3a^2+b^2)(a^2+3b^2)$$Now we show that $a$ and $b$ cannot be the same parity. Clearly, $a$ and $b$ can't both be even since they are relatively prime. If they are both odd, then $a^2 \equiv b^2 \equiv 1 \pmod 8$ so $a^2+3b^2 \equiv 3a^2+b^2 \equiv 4 \pmod 8$. It follows that $v_2(ab(3a^2+b^2)(a^2+3b^2)) = 4$ while $v_2(d(a-b)^7) \geq 7$, contradiction. Now, by Euclidean Algorithm $\gcd(a-b,3a^2+b^2) = \gcd(a-b, 3a^2+b^2-(b^2-a^2)) = \gcd(a-b, 4a^2) = 1$ since $\gcd(a-b,a) = 1$ and since $a-b$ is odd. Similarly, $\gcd(a-b,a^2+3b^2) = \gcd(a-b,a^2+3b^2-(a^2-b^2)) = \gcd(a-b,4b^2) = 1$. Thus if we must have $a-b \mid (3a^2+b^2)(a^2+3b^2)$ then $a-b = 1$. Then we just allow $b = n$ and $a = n+1$ for some integer $n$ which is enough to express $d$ in terms of $n$. Then we simply take $x = d(n+1)$ and $y = dn$ and since $n$ can take on infinitely many values we're done.

Substitute in $a=x+y,b=x-y$ can give us $\dfrac{1}{4}(a^6-b^6)=b^7$. From there we can get $a=\pm b\sqrt[6]{4b+1}$, so the solutions $(x,y)=(\dfrac{1}{2}b(\pm \sqrt[6]{4b+1}+1),\dfrac{1}{2}b(\pm \sqrt[6]{4b+1}-1)$ A integer solution is obtained whenever $4b+1$ is a perfect 6th power. Full proof here: https://infinityintegral.substack.com/p/usajmo-2017-contest-review

So standard it's almost trite. Let $d = \gcd(x, y)$, and let $x = da$, $y = db$, so the problem reduced to $$ab(3a^2+b^2)(a^2+3b^2) = d(a-b)^7.$$ Claim. $a$ and $b$ are of opposite parity. Proof. If they are both odd, then $\nu_2(3a^2+b^2) = \nu_2(a^2+3b^2) = 2$, so we have a $\nu_2$ contradiction. $\blacksquare$ Then $(a-b, 3a^2+b^2) = (a-b, 4a^2) = 1$, so $a-b$ is relatively prime to every term on the LHS. It follows $a-b = 1$, and this sufficiently describes all solutions.

Yet another problem that gets killed by a standard substitution trick. By $\pmod{2}$ inspection, $x, y$ cannot have different parities, so let $x = a + b, y = a - b$. The equation becomes \[(3x^3 + xy^2)(x^2y + 3y^3) = (x - y)^7 \iff xy(3x^2 + y^2)(x^2 + 3y^2) = (x - y)^7 \]\[\iff 16(a + b)(a - b)(a^2 + ab + b^2)(a^2 - ab + b^2) = 128a^7 \iff (a^2 - b^2)(a^4 + a^2b^2 + b^4) = 8a^7 \]\[\iff a^6 - b^6 = 8a^7 \iff a^6(8a + 1) = b^6. \]So $8a + 1$ is a perfect sixth power, which are all determined by some odd $k^6$. Then $a = \dfrac{k^6 - 1}{8}$ and consequently, $b = \dfrac{k^7 - k}{8}$. Solving for $x, y$ yields the curve \[ (x, y) = \left(\dfrac{k^7 + k^6 - k - 1}{8} , \dfrac{k^7 - k^6 - k + 1}{8} \right) \]of solutions for all odd $k$.