literally just do a lotta stuff until u get (2x^2-1)(2y^2-1)(2z^2-1)=2023, at which point it is easy to see only (2,3,3) and perms work

Expand the equation and factor to get $(2x^2-1)(2y^2-1)(2z^2-1) = 2023$ and casework to get $(2,3,3)$ and permutations Should have been a AMC 10 Problem.

did too much busy work with sum of squares to prove that $x \leq 2$ (WLOG $x\leq y \leq z$), and then when checking $x=1,2$, found the factorization. bruh

Alternate solution: Prove $x + y + z + 2xyz \ge 2xy +2yz + 2zx + 1$ for $x,y,z \ge 3$. Then bash the cases when $\min(x,y,z) \in \{1,2\}$. End up with $(2,3,3)$ and permutations.

By Napkin 47.1.5 there is a norm function $\operatorname{Norm} \colon {\mathbb Q}(\sqrt2) \to {\mathbb Q}$ defined by \[ \operatorname{Norm}(a+b\sqrt2) = a^2-2b^2 \]which is multiplicative, meaning \[ \operatorname{Norm}(u \cdot v) = \operatorname{Norm}(u) \cdot \operatorname{Norm}(v). \]This means that for any rational numbers $x$, $y$, $z$, we should have \[ \operatorname{Norm} \left( (1+\sqrt2x)(1+\sqrt2y)(1+\sqrt2z) \right) = \operatorname{Norm}(1+\sqrt2x) \cdot \operatorname{Norm}(1+\sqrt2y) \cdot \operatorname{Norm}(1+\sqrt2z). \]Since $(1+\sqrt2x)(1+\sqrt2y)(1+\sqrt2z) = (2xy+2yz+2zx+1) + (x+y+z+2xyz)\sqrt2$, this gives the identity \[ (2xy+2yz+2zx+1)^2-2(x+y+z+2xyz)^2 = (1-2x^2)(1-2y^2)(1-2z^2) \]which kills the problem.

$(\sqrt2(x+y+z)+2\sqrt2xyz)^2-(2xy+2yz+2zx+1)^2=2023$ implies $(\sqrt2x+1)(\sqrt2y+1)(\sqrt2z+1)(\sqrt2x-1)(\sqrt2y-1)(\sqrt2z-1)=2023$, so $(2x^2-1)(2y^2-1)(2z^2-1)=2023$

Just expand and factor... After cancel and everything, $(2x^2-1)(2y^2-1)(2z^2-1) = 2023$ And we know $2013 = 17*17*7$, so just $(2,3,3)$ and permutations imo this is not as hard as i would expect a jmo

$50 and two icecream cakes Titu wrote this

so how many points if, while doing algebraic manipulation, i wrote the wrong constant, which didnt affect the factoring, just the last step

I didn't solve this problem; I tried expanding, but forgot about the 2 coefficient in the LHS when doing so, and got a big mess.

The motivation for moving the $2$ inside is actually pretty simple: things like $(2x+1)(2y+1)(2z+1)$ usually have geometric sequences as coefficients and so you *really* want that to happen here implying the result

Expand then set $a=2x^2, b=2y^2, c=2z^2.$ You get $a+b+c-(ab+bc+ca)+abc=2024$ which is much like a polynomial f(1) with roots $a, b, c.$ This is the motivation then do the stuff mentioned above

for me i tried some false bounding claims that were proven with faulty calcbash. then i realized "wait why am i using calc on a jmo 1" and then i found the factorization after looking at some of the expressions on my scratch paper.

We present an elegant solution to this difficult problem: WLOG $x\ge y\ge z$. Expanding and simplifying, we have the equation \[x^2+y^2+z^2+4x^2y^2z^2=2x^2y^2+2y^2z^2+2x^2z^2+1012\quad(*)\]Suppose $z=1$. Then, the equation simplifies to \begin{align*}& x^2+y^2+1+4x^2y^2=2x^2+2y^2+2x^2y^2+1012 \\&\implies (2x^2-1)(2y^2-1)=2023=17^2\cdot 7 \end{align*}Since $x\ge y$, we see $2x^2-1=7\cdot 17$, $17^2$, $7\cdot 17^2$, which do not yield integer solutions for $x$. Then, $x,y,z\ge 2$ so \begin{align*}(\tfrac{1}{2}x^2)y^2z^2&\ge 2y^2z^2\\ (\tfrac{1}{2}y^2)x^2z^2&\ge 2x^2z^2\\ (\tfrac{1}{2}z^2)x^2y^2&\ge 2x^2y^2\end{align*}so \[\tfrac{3}{2}x^2y^2z^2\ge 2x^2y^2+2y^2z^2+2z^2x^2\]Then, by $(*)$, we have \begin{align*}\frac{5}{2}\cdot 2^2\cdot 2^2\cdot x^2+x^2&\le \tfrac{5}{2}x^2y^2z^2+x^2+y^2+z^2\\&=2x^2y^2+2y^2z^2+2z^2x^2-\tfrac{3}{2}x^2y^2z^2+1012\\&\le 1012\end{align*}Hence, $41x^2\le 1012$ so $x<5$. Notice taking $(*)$ modulo $2$ that we have $x^2+y^2+z^2\equiv 0\pmod{2}$ so all of $x$, $y$, $z$, are even or exactly one of them are even. We know $2\le x,y,z\le 4$. Case 1: $x=2$. Then $y=z=2$. Plugging into $(*)$, we see this does not work. Case 2: $x=3$. Then, by parity, $y=3$ and $z=2$. Plugging this in, we see that it works. Case 3: $x=4$.Then, $(y,z)=(4,4),(4,2),(2,2),(3,3)$. Plugging all of these in we see that they all fail. Hence, our only solutions are $(x,y,z)=(2,3,3),(3,2,3),(3,3,2)$.

brug :skull: at least you did it

oh my god im getting a 6 on this i didnt see the elegant factorization so instead i used rearrangement to bound (got xy+yz+zx<=69 nice) then i bashed out every single case

mikimoto12 wrote: oh my god im getting a 6 on this i didnt see the elegant factorization so instead i used rearrangement to bound (got xy+yz+zx<=69 nice) then i bashed out every single case if its right then it should still be a 7

Note the miraculous equality \[ (1 + x\sqrt{2})(1 + y\sqrt{2})(1 + z\sqrt{2}) = 1 + 2xy + 2yz + 2zx + \sqrt{2} \left(x + y + z + 2xyz \right) \]by which we can consider the norm function $N:\mathbb Z[\sqrt{2}]\to \mathbb Z$ \[ N(a + b\sqrt{2}) = a^2 - 2b^2. \]Particularly, we have the fact that for any $p, q \in \mathbb Z[\sqrt{2}]$, we find $N(p)N(q) = N(pq)$. This is left for the reader to think about (check by direct expansion, similar to how complex conjugates work... hmm...) As such, \[ 2023 = -N(1+x\sqrt{2})N(1+y\sqrt{2})N(1+z\sqrt{2})\]or in other words, \[ 2023 = (2x^2 - 1)(2y^2 - 1)(2z^2 - 1)\]which leads to the default solutions of $(2, 3, 3)$ and cyclic. $\textbf{Remark.}$ This is extremely motivated given that you know a tiny amount of algebraic NT. (shoutouts to PROMYS to introducing some of this) The point is to notice the rather SFFT-like expressions in $1 + 2xy + 2yz + 2x$ and $x + y + z + 2xyz$, and then realize that the $2$ exists, which naturally gives something of the form $p^2 - 2q^2$; it naturally comes to mind that $(1 + x\sqrt{2})(1 + y\sqrt{2})(1 + z\sqrt{2})$ will do the trick. As per the whole multiplicativity thing, I will leave it to the reader to consider stuff like $|a + bi|^2$, where the norm function now is $a^2 + b^2$. Similar stuff like \[ 5(x + y + z + w + 5(xyz + yzw + zwy + wyx))^2 = (25xyzw + 5(xy + yz + zw + wx + xz + yw) +1)^2 - 301796 \]can be nuked in the exact same way. From there it's just off to the computation. Imagine expanding, :3

POV: 3/7 $$2024= 2(x+y+z+2xyz)^2-(2xy+2yz+2zx+1)^2+1 = 8 x^2 y^2 z^2 - 4 x^2 y^2 - 4 x^2 z^2 + 2 x^2 - 4 y^2 z^2 + 2x^2+2 y^2 + 2 z^2$$. Divide by $2$ on my scratch paper to get $$1012 = 4x^2y^2z^2-2x^2y^2-2y^2z^2-2x^2z^2+x^2+y^2+z^2$$Realize that you can factor so the problem is trivial, go do another problem, come back for write-up. Looking at the scratch paper, write $$1013 = (2x^2-1)(2y^2-1)(2z^2-1)$$Get no solutions. When I receive my score distribution, see that i missed MOP by... 4 points

havent done this much expansion in a while :O Of course, we expand the given equation, and some pain has been spared to give us \[8x^2y^2z^2-4x^2y^2-4y^2z^2-4z^2x^2+2x^2+2y^2+2z^2=2024\]which factors as \[(2x^2-1)(2y^2-1)(2z^2-1)=2023=7\cdot 17^2.\]Casework.. It's easy to see that the possible triples of $(2x^2-1, 2y^2-1, 2z^2-1)$ are $(7, 17, 17)$, $(117, 17, 1)$, $(2023, 1, 1)$, $(289, 7, 1)$, $(2023, -1, -1)$, and permutations. We discard all but $(7, 17, 17)$ because they don't have integral solutions :p If the $2x^2-1$ and similar terms are $(7, 17, 17)$ in some order, then $(x, y, z)$ is a permutation of $(2, 3, 3)$. So that's the answer tada~

Expanding and simplifying: $2x^2+2y^2+2z^2+8x^2y^2z^2-4x^2y^2-4y^2z^2-4z^2x^2-1=2023$ Factorising: $(2x^2-1)(2y^2-1)(2z^2-1)=2023$ Prime factorisation of 2023: $17^2*7$ So $2x^2-1=17$ $2x^2=18$ $x^2=9$ $x=3$ Doing the same thing will give $y=3,z=2$ so only $(3,3,2)$ as well as permuatations will work

Okay now I wonder why I never saw norms Let $\mathcal{N}(a\sqrt{2}+b)=2a^2-b^2$ denote the norm in $\mathbb{Z}[\sqrt{2}]$, and note its multiplicative nature. Move $(2xy+2yz+2zx+1)^2$ to the LHS of the equation. The resulting LHS of rearranges as \[ \mathcal{N}((2xyz+x+y+z)\sqrt{2}+2xy+2yz+2xz+1) = \mathcal{N}((x\sqrt{2}+1)(y\sqrt{2}+1)(z\sqrt{2}+1)) = (2x^2-1)(2y^2-1)(2z^2-1). \]Thus, we need to find the number of positive integer solutions to $(2x^2-1)(2y^2-1)(2z^2-1)=2023$. The only solutions are $(2, 2, 3)$ and its permutations, and we conclude.

RedFireTruck wrote: literally just do a lotta stuff until u get (2x^2-1)(2y^2-1)(2z^2-1)=2023, at which point it is easy to see only (2,3,3) and perms work my goofy ahh ended up getting a 6/7

did you get deducted for the last part? "notice that the only values of x such that 2x^2-1 | 2023 are x=1,2,3." I didn't really see anything wrong so I was wondering what the deduction was for

He didn't specify that they satisfy the equation. @3below It probably would be a 7 if he specified that they satisfied the equation. Although the steps are reversible, he didn't say that $(x,y,z) = (2,2,3)$ satisfied $(2x^2 - 1)(2y^2 - 1) (2z^2 - 1) = 2023$ anyway.

So you need to show (2,3,3) and permutations work? or the x=1,2,3 work?

LearnMath_105 wrote: So you need to show (2,3,3) and permutations work? or the x=1,2,3 work? You need to show that the find $(2,2,3)$ and permutations actually work. What he did was just proving that: if $x,y,z$ satisfy the equation, then $(x,y,z) = (2,2,3)$ (and permutations). He never showed that $(x,y,z) = (2,2,3)$ implies that $x,y,z$ satisfy the equation. In other words, he only showed one direction of the if and only if that the problem wanted. Page 8 of https://web.evanchen.cc/textbooks/OTIS-Excerpts.pdf#page17 has more about this detail.

@naonaoaz AoPS User still would prob get a 6 if this was actual jmo and plus i think all steps are reversible anways so that is not necessary

Oof I wish all JMO #1's were like this

We can write $2023=-1+2x^2+2y^2+2z^2-4x^2y^2-4y^2z^2-4z^2x^2+8x^2y^2z^2$, which we can easily factor as $(2x^2-1)(2y^2-1)(2z^2-1)$. Now it's easy to get $(x,y,z)$ is a permutation of $(2,3,3)$.

After expanding and simplifying the LHS and RHS calmly, we get 8(xyz)² + 2x² + 2y² + 2z² - 4x²y² - 4y²z² - 4z²x² - 1 = 2023 (2x² - 1)(4y²z² - 2y² - 2z² + 1) = 2023 (2x² - 1)(2y² - 1)(2z² - 1) = 2023 = 7.17² v_7(2x² - 1)(2y² - 1)(2z² - 1) = v_7(2023) = 1 This implies that one of 3 factors present in the LHS is equal to 7. v_17(2x² - 1)(2y² - 1)(2z² - 1) = v_17(2023) = 2 Similarly, this implies exactly 2 of the 3 factors present in the LHS are equal to 17. Now, it's easy to get (x,y,z) = (2,3,3),(3,2,3) and (3,3,2).