$q=0$ is apparently the solution by the case $a:=e-f$, $b:=g-h$, $c:=f-e$, $d:=h-g$ where $e,f,g,h$ are the numbers on the napkin but from then... it'd be a tricky work

q = 2 and -2 also work

Claim 1: $2 \ge |q|$ Proof : Our 10 distinct numbers are $a_1,a_2,...,a_{10}$, where $a_{10} \ge a_9 \ge ...\ge a_1$. Then maximum of second line is $q(a_{10}-a_1)^2$ and maximum of third line is $2(a_{10}-a_1)^2$, where $q \ge 0$. We get $2 \ge q \ge 0$. Also we know if $M$ is in third line. Then $-M$ is also in third line. So $q$ is solution,then $-q$ is solution.And $q$ is not solution, the $-q$ is not solution. So $2 \ge |q|$. Claim 2 : q is an integer Proof : If ten distinct numbers are $1,b_1,...,b_9$, where $ b_i $ (i=1,2,...,9) is an integer. Then we get q is an integer. So $q=2,1,0,-1,-2$. Case 1: $q=0$ Then second line is only zero.And we know in the third line we have zero.($a^2+a^2-a^2-a^2$). Case 2 : $q=2,-2$ Construction: If $2ab$ or $-2ab$ in the second line, then a and b in the first line . This means $a+x,x,b+y,y$ in the first 10 distinct numbers, where x and y is different and x and y is not zero.This means $a+x-b-y,a+x-y,b+y-x,x-y$ is in the first line. So $2ab=(a+x-y)^2+(b+y-x)^2-(a+x-b-y)^2-(x-y)^2$ and $-2ab=(a+x-b-y)^2+(x-y)^2-(a+x-y)^2-(a+y-x)^2$ Case 3: $q=1,-1$ 1,2,5,9,13,17,21,25,29,33 be 10 distinct numbers.All numbers from first line $\equiv 0,1,3,4,-3,-1$ (mod 8). Exist number $a$ in the second line, $a \equiv 4 $(mod 8).All numbers from third line $\equiv 0,1,-1,2,-2$ (mod 8). But in the secod line we have number such that $\equiv 4 $(mod 8). Solution:{0,2,-2}

We claim that $\boxed{q\in \{-2,0,2\}}$ is the complete solution set. Note that $q=0$ trivially works since $0\in S,$ and further note that since $a^2+b^2-c^2-d^2=-(c^2+d^2-a^2-b^2),$ $q$ being a solution implies that $-q$ is a solution. Thus, from now on, we may assume that $q$ is positive. Taking $S=\{0,1,\cdots, n-1\},$ we see that $q=q\cdot 1\cdot 1\in B\implies q\in C,$ so $q$ must be an integer. We now break into cases: Case 1: $q=1.$ Take $S=\{0,1,\cdots, n-2\}\cup\{\sqrt 2\}.$ Then $\sqrt2 = q\cdot (\sqrt2 - 0)\cdot (1-0)\in b\implies \sqrt2 \in C.$ But note that for each $a\in A,$ $a=b+c\sqrt2$ for some integers $b,c,$ implying $a^2=(b+c\sqrt 2)^2=b^2+2c^2+2bc\sqrt2.$ But the coefficient of $\sqrt2$ here is always even, which is a contradiction since $1,\sqrt2$ are linearly indepedent. Case 2: $q=2.$ This indeed works, as seen from the identity \[2(a-b)(c-d)=(a-d)^2+(b-c)^2-(a-c)^2-(b-d)^2.\] Case 3: $q>2.$ Let $x=\text{max}(A).$ Then $qx^2\in C,$ but \[qx^2>2x^2= (x^2+x^2+0+0)\geq a^2+b^2-c^2-d^2,\]contradiction. Hence, we conclude with the solutions stated at the beginning.

First consider the case where the napkin contains $\{0, 1, \dots, 9\}$. The set of differences is $\{-9, -8, \dots, 8, 9\}$. Notice then that for any $a, b, c, d$ in this set we have $\vert a^2 + b^2 - c^2 - d^2 \vert \leq 162$. Moreover we have that $81q$ and $-81q$ both belong to the second line, which implies due to the previous inequality that $\vert q \vert \leq 2$. Moreover, $q = q \cdot 1 \cdot 1$ is in the second line, which implies that $q$ is an integer, as every number in the set of differences is an integer and thus every number in the third line is an integer too. This implies $q \in \{-2, -1, 0, 1, 2\}$. We now show that $q = 1$ and $q = -1$ are impossible. Consider the case when the napkin contains $\{0, 1, 5, 9, \dots, 33\}$. The differences in this set are all either odd or multiples of $4$, so their squares are all $0$ or $1 \bmod{8}$. This implies that $a^2 + b^2 - c^2 - d^2$ is in $\{-2, -1, 0, 1, 2\} \bmod{8}$. Because $1$ and $4$ belong to the difference set, we find that $1 \cdot 1 \cdot 4 = 4$ is written on the second line, but no number on the third line is congruent to $4 \bmod{8}$, so the condition isn't satisfied, and thus $q = 1$ is impossible. The case $q = -1$ is analogous. Finally, by noticing the following identities: $$(a-b)^2 + (a-b)^2 - (a-b)^2 - (a-b)^2 = 0$$$$(a - d)^2 + (b - c)^2 - (a - c)^2 - (b - d)^2 = 2(a - b)(c - d)$$$$(a - c)^2 + (b - d)^2 - (a - d)^2 - (b - c)^2 = -2(a - b)(c - d)$$ We find that $q \in \{-2, 0, 2\}$ does work.

Sorry for the non-mathematical reply, but I'd like to point out that Gugu (Carlos Gustavo Moreira) is a former olympian (gold medallist in IMO '90) and now mathematician here from Brazil. He is a very dear member of the Math Olympiad community and has the habit of scribbling math problems and solutions in napkins whenever we go to a restaurant. So the Gugu character is sort of an internal joke.

We claim only $q=-2,0,2$ work. Clearly $q=0$ works. Now we show $q=\pm 2$ works. Suppose the number $\pm 2(a-b)(c-d)$ is on the second line where $a,b,c,d$ are one of the original $10$. Then, the number $\pm((a-d)^2+(b-c)^2-(a-c)^2-(b-d)^2) = \pm 2(ac+bd-ad-bc)=\pm 2((a-b)(c-d))$ is on the third line, as desired. Now we show that $q\in\{-2,0,2\}$. Suppose the first ten numbers were one $1$ and nine $0$s. Then, the first line would be a bunch of $\{1,0,-1\}$, so the second line would be a bunch of $\{-q,0,q\}$, and the third line is a bunch of $\{2,1,0,-1,-2\}$, so $q\in\{-2,-1,0,1,2\}$. It suffices to show now that $q=\pm 1$ doesn't work (we'll just do $q=1$, the other follows trivially). I couldn't find a neat example, but here's one that works. Suppose we start with a bunch of $0,3,5$. Then the first line is a bunch of $\pm\{0,2,3,5\}$. The second line has $6$. We show the third line does not. The set of all $\{a^2+b^2\}$ is $\{0,4,9,25,8,18,50,13,29,34\}$. We can check that no two of these differ to $6$, as desired. Therefore, $\boxed{q=-2,0,2}$. Remark: The first thing I did was the reduction to $q=-2,-1,0,1,2$. Then I quickly saw that $q=-2,0,2$ work, and there is no way on this earth that $q=1$ was going to work in general. The case I used to show this was the first thing that my pen wrote on the paper.

Bashy problem. Anyway, here's my solution: We start off with the following Lemma. Lemma $q$ is an integer such that $-2 \leq q \leq 2$. PROOF: Suppose that the numbers on Gugu's napkin are $\{-4,-3,-2,-1,0,1,2,3,4,5\}$. Then the first line has integers ranging from $-9$ to $9$. As the first line has only integers, the third line must also contain integers. This gives that $q$ is also an integer (as $q$ is a part of the second line for $a=b=1$). Now, taking $a=b=9$ and $a=b=-9$ in the second line, we have that $81q$ and $-81q$ are a part of the second line (These are the largest multiples of $q$ in the second line). Notice that the maximum possible value in the third line is $162$ for $a=\pm 9,b=\pm 9,c=0,d=0$. Similarly, the smallest possible value in the third line is $-162$ for $a=0,b=0,c=\pm 9,d=\pm 9$. Thus, we get that $|81q| \leq 162 \Rightarrow |q| \leq 2 \Rightarrow -2 \leq q \leq 2$. $\Box$ Return to the problem at hand. With a heavy heart, let's begin the bashing . CASE-1 $(q=0):$ This means that the second line has only zero. Taking $a=b=c=d$, we have that zero is a part of the third line also. Thus, $q=0$ works. CASE-2 $(q=\pm 2):$ We claim that $q=\pm 2$ works. This claim is direct from the following identity, which is true $\forall a,b,c,d \in \mathbb{R}$- $$(a-b)^2+(c-d)^2-(b-c)^2-(d-a)^2=2(bc+da-ab-cd)=2(a-c)(d-b)$$ CASE-3 $(q=\pm 1):$ We claim that $q=\pm 1$ doesn't work. For this purpose, suppose that the initial numbers on Gugu's napkin are $\{2m,2m+1,2m+5, \dots, 2m+33\}$ for some $m \in \mathbb{Z}$. Then the first line has numbers $\equiv 0$ or $\pm 1 \pmod{4}$. This gives that the second line (i.e. $ab$) has numbers $\equiv 0$ or $\pm 1 \pmod{4}$. Note that the squares of the numbers in the first line are either $0$ or $1 \pmod{8}$. This means that the third line (i.e. $a^2+b^2-c^2-d^2$) has numbers $\equiv 0$ or $1 \pmod{8}$. But then all those numbers lying in the second line, which are $\equiv 4 \pmod{8}$ (such as the product of $(2m+1)-(2m)=1$ and $(2m+5)-(2m+1)=4$) are no longer a part of the third line. Thus, Our claim is true. Thus, The final answer is $q \in \{-2,0,2\}$.

yayups wrote: We claim only $q=-2,0,2$ work. Clearly $q=0$ works. Now we show $q=\pm 2$ works. Suppose the number $\pm 2(a-b)(c-d)$ is on the second line where $a,b,c,d$ are one of the original $10$. Then, the number $\pm((a-d)^2+(b-c)^2-(a-c)^2-(b-d)^2) = \pm 2(ac+bd-ad-bc)=\pm 2((a-b)(c-d))$ is on the third line, as desired. Now we show that $q\in\{-2,0,2\}$. Suppose the first ten numbers were one $1$ and nine $0$s. Then, the first line would be a bunch of $\{1,0,-1\}$, so the second line would be a bunch of $\{-q,0,q\}$, and the third line is a bunch of $\{2,1,0,-1,-2\}$, so $q\in\{-2,-1,0,1,2\}$. It suffices to show now that $q=\pm 1$ doesn't work (we'll just do $q=1$, the other follows trivially). I couldn't find a neat example, but here's one that works. Suppose we start with a bunch of $0,3,5$. Then the first line is a bunch of $\pm\{0,2,3,5\}$. The second line has $6$. We show the third line does not. The set of all $\{a^2+b^2\}$ is $\{0,4,9,25,8,18,50,13,29,34\}$. We can check that no two of these differ to $6$, as desired. This solution doesn't work, as you have taken some of the numbers on the napkins to be same, even though they are given to be distinct.

Wait but like the problem says $a$ and $b$ can be the same. I guess I interpreted the wording differently than what it was, but I think I am sort of right in saying that the ideas are the same, just that with the ``correct'' wording, you just have to be a lot more careful.

yayups wrote: Wait but like the problem says $a$ and $b$ can be the same. I guess I interpreted the wording differently than what it was, but I think I am sort of right in saying that the ideas are the same, just that with the ``correct'' wording, you just have to be a lot more careful. I think math_pi_rate’s complaint is that the problem explicitly states that the ten initial numbers are all different (while you consider e.g. one 1 and nine 0s). Of course you can fix it by making the other nine numbers small in magnitude but all different.

CantonMathGuy wrote: Of course you can fix it by making the other nine numbers small in magnitude but all different. Sorry but I don't really get how this would help. Cause then the numbers in the first line will no longer be integers, and the problem will become even harder.

Is there a less bashy way to solve this problem?

Here’s my solution CLAIM 1: $q$ is an integer. PROOF: If the ten numbers which were considered at the beginning were integers, then it is necessary that $q$ is an integer. CLAIM 2: $|q| \leq 2$. PROOF: Order the $10$ reals in ascending order. Let them be $a_1 \leq a_2 \leq a_3 \leq a_4 .... \leq a_10.$ Max. number in second line is $q(a_{10}-a_1)^2 $;while the Max. number in third line is $2(a_{10}-a_1)^2.$ Hence the desired conclusion follows. CLAIM 3: Only $q = 2,0,-2$ satisfy the required condition. PROOF : It is easy to see the above values satisfy the given condition. Suppose $q = 1$. Consider the set ${0,1,5,9,13,....,33}.$ The numbers in the second line are always $1,0\pmod 4$. But all the numbers in the third line are $0\pmod 4$. Contradiction. A similar argument holds true for $q = -1$ Hence required condition follows. So, $\boxed { q = 2,0,-2 }$

Meh problem... just based on $(a-c)^2+(b-d)^2-(a-d)^2-(b-c)^2=2(ad+bc--ac-bd)=2(a-b)(d-c)$, hence $q=+-2$ work, and trivially $0$ works as well. Putting $a,b,0,0,...,0$ gives a gist for why these are the only $q$ that work, and if we let 8 of the numbers to be as small as we want, then we only need to check if $q(a^2-ab)$ is achievable by $a,b,a-b,0$, and as $q$ is an integer (set all 10 numbers to integers), we get that only $+-2$ is possible.

Let $P(a_{1}, a_{2}, ..., a_{10})$ be the $10$ real numbers. $P(0, 0, 0, 0, 0, 1, 1, 1, 1, 1)$ (or any number of $0$'s and $1$'s with at least one of each) implies $q \in \{-2, -1, 0, 1, 2\}$. $q = 0$ is trivial (as letting $a = b = c = d$ in $a^2 + b^2 - c^2 - d^2$ gives you a zero in the third row). Let's notice that if $x$ is a solution for $q$, then so is $-x$ ($\because$ if $xmn = a^2 + b^2 - c^2 - d^2$ for some $a, b, c, d \in \{a_{1}, ..., a_{10} \}$, where no two of $a, b, c,$ and $d$ are necessarily distinct, then $-xmn = c^2 + d^2 - a^2 - b^2$). This implies that is sufficient to check whether $q = 1$ and $q = 2$ work. $1^{\circ}$ $q = 1$ Plugging in $P( 0, ..., 0, 3, 4)$ implies that the numbers in the first line will be $-4, -3, -1, 0, 1, 3, 4$, meaning we will have $3 \cdot 4 = 12$ in the second line, however we won't have it in the third line, as $a^2 + b^2 - c^2 - d^2$ can never equal $12$ for $a, b, c, d \in \{-4, -3, -1, 0, 1, 3, 4\}$, i.e. $a, b, c, d \in \{0, 1, 3, 4\}$ (this can be easily proven by concluding that $a^2 + b^2 - c^2 - d^2 > 10$ iff $|a|, |b| \ge 3$, and that $a^2 - b^2$ cannot be equal to $12$ within the set of possible values for $a$ and $b$, therefore $|a| \neq |c|, |d|$ and $|b| \neq |c|, |d|$, so we have the $3$ easy cases to check) Similarly, $q = -1$ also doesn't work. $2^{\circ}$ q = 2 $2(a_i - a_j)(a_k - a_l) = (a_i - a_l)^2 + (a_j - a_k)^2 - (a_i - a_k)^2 - (a_j - a_l)^2$, which implies that all numbers in the second line are also present in the third line. Therefore, both $q = 2$ and $q = -2$ work. Finally, $q \in \{-2, 0, 2\}$ is the set of solutions.

The answer is $q=2, 0, -2$ $q=0$ is easy, assume that $q \neq 0$. Let the $10$ number be $a_1\le a_2\le \cdots \le a_n$, call it set $A$. Define $$g(a, b, c, d)=(a-b)(c-d) \newline$$$$ f(a, b, c, d)=(a-d)^2+(b-c)^2-(a-c)^2-(b-d)^2$$$$ X_q=\{ g(a, b, c, d) |a, b, c, d \in A\}$$$$Y=\{ f(a, b, c, d)|a, b, c, d, \in A\}$$$X_q, Y$ is the number of the second line and the third line. Notice that $$ f(a, b, c, d) = (a-d)^2+(b-c)^2-(a-c)^2-(b-d)^2 =2(a-b)(c-d)= g(a, b, c, d) $$hence $Y=X_2$. Since $|X_q|=|X_2|$ for all $q\neq 0$, so that $X_q \subset Y=X_2$ if and only if $X_q=X_2$. Look at the range of the set, hence $q=\pm 2$.

Claim: $q$ is an integer. proof.At first observe that $q$ can't be a irrational.Otherwise,consedering 10 integers as Gugu's initial list,there is a irrational number,namely $q\times ab$ in the second line where $a,b\in\mathbb Z$.On the other hand,every number in the 3rd line is an integer. So $q$ is a rational.Consider the initial list so that it produces atleast 9 primes and everything integer in the 1st list.For example,$\{x,x+p_1,x+p_2,x+p_3\dots x+p_9\}$ where$p_i$'s are prime and $x\in\mathbb Z$. Then comparing with 3rd line $q\times p_1p_2$ is an integer and $q\times p_3p_4$ is an integer.Hence, denominator of $q$ must be 1.Hence $q\in \mathbb Z$.$\blacksquare$ Now observe that if $q$ works then so is $-q$ as we can take $c^2+d^2-a^2-b^2$ this time insteed of $a^2+b^2-c^2-d^2$.Also note that $q=0$ is a solution.Hence,we'll only try to find positive solution. Claim:$ q$ can not be greater than or equal to 3 proof.Suppose,$q\ge 3$.Take Gugu's list such a way that the 2 maximum possible differences $x\ge y$ are $\ge 2020^{2020}$ and $2y^2>x^2$.Then $q\times xy\ge 3\times xy\ge 3y^2>x^2+y^2>|a^2+b^2-c^2-d^2|$.$\blacksquare$ Claim: $q\ne 1$ proof.Suppose $q=1$.Consider the list $\{1,2,3,4,5,6,7,8,9,1+e\}$ where $e$ is Euler's number. Then $e$ must be in 2nd linet.For example $(1+e-1)(2-1)$. Now,if $a^2+b^2-c^2-d^2=e$ for some $a,b,c,d$ in the first line then some of $a,b,c,d$ must be of form $(e-x)$ or $(x-e)$ for some $x$ in initial line.But it's square namely,$ x^2-2xe+e^2$ contains even number of $e$'s.Hence there is no $e$ in the third line.$\blacksquare$ Claim: $q=2$ works. proof.Observe that, $2(m-n)(x-y)\\ =2mx-2my-2nx+2ny\\ =(m-y)^2+(n-x)^2-(n-y)^2-(m-x)^2$. Hence, $-2$ also works.$\blacksquare$ Hence,$q=2,-2,0$.$\blacksquare$

first let the $10$ numbers be $$1,2,3,4,5,6,7,8,m+1,n+1$$Where $n,m$ are natural numbers so clearly $q$ is rational . Also it is clear that if $q$ works , then so does $-q$, so we just look for possitive rational $q$ which work ($0$ clearly works , and we dont really care) Assume $q=\frac{i}{j}$ such that $(i,j)=1$ we prove $j=1$ Assume this is not the case . from here let $A_1,A_2,A_3$ denote the set of first , the second and the third row of the written numbers(for example $1,n \in A_1$) clearly all numbers in $A_3$ are integers , then so should be the numbers in $A_2$ , assume $m=j+n-1$ so we get $$\frac{i(j-1)^2}{j} \in A_2$$but clearly is not an integer , so we must have $j=1$. Which gives $q$ is an integer.let $m=9$ so we have the number $$qn^2=a^2+b^2-c^2-d^2$$for some $a,b,c,d \in A$ but clearly if $a \in A$ then $|a| \le n$ so we have $$qn^2 \le a^2+b^2 \le 2n^2$$so $q \le 2$ Clearly $q=2$ works fine , so we just need to check $q=1$ for this take the set $$(1,2,3,4,5,6,7,8,9,1+2^{\frac{1}{3}})$$we have $$2^{\frac{1}{3}}(2^{\frac{1}{3}}-1)\in A_2$$Let $2=t^3$ hence we have $$t(t-1)=a^2+b^2-c^2-d^2$$now let $a=i+jt, b=u+vt , c=n+mt, d=o+kt$ Where $j,v,m,k \in \{-1,0,1\}$ and $i,u,n,o$ are done natural numbers so we have $$t^2-t=t^2(j^2+v^2-m^2-k^2)+2t(ij+uv-nm-ok)+(i^2+u^2-n^2-o^2)$$but since $1,t,t^2$ are linerly independant we have $$-1=2ij+2uv-2nm-2ok$$which is ocntradiction since one side is odd and the other is even. so $ q \in \{-2,0,2\}$

silly problem; with awang11 The answer is $q=0, \pm 2$. Let Gugu's numbers be $a_1, \dots , a_{10}$. First we demonstrate all other $q$ are bad; viewing everything Gugu writes as a multivariable polynomial, if two terms match infinitely many times, they must be identical polynomials. Hence considering the term $$q(a_1-a_2)(a_1-a_3)=qa_1^2-q(a_2+a_3-a_2a_3)$$the coefficient of the $a_1^2$ term is $q$, and since this polynomial must appear in the third line, it's clear $q$ must be an integer with $|q| \le 2$. In addition, $q= \pm 1$ fails since it's easy to see that for any polynomial in the third line, the sum of the coefficients of terms of the form $a_i^2$ must be even. Finally we verify $q=0, \pm 2$ indeed works; when $q=0$ this is obvious, and when $q=\pm 2$ we have the following identity for not necessarily distinct $i,j,k,n$: $$\pm 2(a_i-a_j)(a_k-a_n) = \pm ( (a_i-a_n)^2+(a_j-a_k)^2-(a_i-a_k)^2-(a_j-a_n)^2)$$the end.

dame dame

The answer is $q=-2,0,2.$ Note that it is obvious that $q$ works if and only if $-q$ works. Claim: $-2\le q\le 2.$ Proof. It suffices to assume $q$ nonnegative and show that $q\le 2.$ Indeed, suppose that the largest number in the first line is $A >0.$ Then, $qA^2$ must be in the third line, which has largest number $2A^2,$ giving $q\le 2.$ $\blacksquare$ Claim: $q$ is an integer. Proof. It is obvious by considering the 10 numbers $0,1,2,\dots, 9$ that all numbers in the third line are integers, implying $q$ must be rational. Suppose $q = m/n$ for $\gcd(m,n) = 1.$ Then $2m/n$ and $3m/n$ must be integers, implying $n\mid m$ and consequently $n=1.$ $\blacksquare$ Claim: $q=2,0,-2$ work and $q=1,-1$ fail. Proof. Obviously $q=0$ works. For $q=2,$ if 4 of the original ten numbes are $w,x,y,z$ and $a=y-x$ and $b=z-w,$ then $$2ab = 2(y-z)(z-w) = (z-x)^2+(w-y)^2 - (x-w)^2-(y-z)^2.$$Implying $q=2$ works for any 10 reals and moreover $q=-2$ works as well. For $q=1,$ consider the ten numbers $\pi, 0,1,2,\dots, 8.$ Suppose $\pi$ was in the third and second line. Then, it must be true that for the expression of $\pi$ in the third line, all $\pi^2$ terms cancelled, leaving $$\pi = 2\pi N + k \iff \pi = \frac{k}{1-2N},$$For some integers $N$ and $k,$ where obviously then $1-2N\ne 0,$ implying $\pi\in \mathbb Q,$ absurd. Similarly, $q=-1$ also fails. $\blacksquare$

I mostly disliked this problem but I shall as always post a solution for storage. I was kind of amused after I got that $|q| \leq 2$ and that $q$ must be an integer basically immediately. The construction for $2$ is not too bad, I guess. $\textbf{Claim:}$ $q \in \{0, \pm 1 \pm 2\}$ $\textbf{Proof)}$ Notice that if $|q|> 2$, then if we take the maximum in Line 2 to be $A$, then $qA^2$ has to be contained in Line 3, yet the maximum of Line 3 is $2A^2 < qA^2$ which is a contradiction. Moreover, taking $1,..,10$ to be the numbers on the napkins, $q$ is in Line 2, on the other hand the only numbers in Line 3 are integers meaning that $q$ must be an integer such that $|q| \leq 2$. $\square$ $\textbf{Claim:}$ $q = \pm 1$ doesn't work. Take $1,2,3,..,8,9,9+\sqrt{2}$ and notice that $\sqrt{2}$ is in Line 2 (either by taking $a = \sqrt{2}$ and $b =q$), yet, each number in Line 3 has an even coefficient of $\sqrt{2}$ in $\mathbb{Q}(\sqrt{2})$ which is a contradiction. $\square$ $\textbf{Claim:}$ $q \in \{0,\pm 2\}$ works. $\textbf{Proof)}$ If $q=0$, the only element of Line 2 is $0$ which is also clearly in Line 3. We shall prove that $q=2$ works which clearly implies that $q=-2$ works by switching $(a,b)$ and $(c,d)$ in Line 3. Take $a,b,c,d$ as four numbers on the napkins. Then $$(a-b)^2+(c-d)^2-(b-c)^2-(d-a)^2 = 2(a-c)(b-d)$$meaning that any number in Line 2 is also in Line 3. Consequently, $q = 2$ also works. $\square$ We get that the only reals $q$ satisfying the condition are $0, \pm 2$. $\blacksquare$

Am I the only one who didn't hate this problem? Let $S$ be the set of numbers on the napkin First, pick some numbers, say $S = \{0,1,2,\cdots,9\}$ to see that $q$ must be an integer. Also, pick $a = b = 9$ in line $2$ to see that $|q| \le 2$ Now, $q = 0$ obviously works since pick $a = c, b = d$ in line $3$ Obviously, if $q$ works, so does $-q$, so I will show that $2$ works but $1$ doesn't. To see $2$ works, say $2ab = 2(p-q)(r-s)$ is the number on line $2$ with $p,q,r,s \in S$ Then, observe that $(p+r)^2 + (q+s)^2 - (p+s)^2 - (q+r)^2 = 2(p-q)(r-s)$ and so it appears on the third line as well. Now, to see $q = 1$ fails, choose $S = \{0,1,2,3,4,5,6,7,8,10000\}$ The nonnegative numbers on line $1$ are $0,1,\cdots,8$ and $9992, 9993, \cdots, 10000$, call the first few numbers small and the others big I claim that $N = 9999 \times 10000 = 99990000$ (which appears on line $2$) does not appear in line $3$. Suppose it did. Then, see that at least one of $a,b$ has to be big Suppose $a > b, c > d$. Note that $a,b,c$ must be big and $d$ small since otherwise only $a$ is big, but its square is sufficiently far from $N$ so it needs a big number for that. Let $M = 10000$ for convenience so $N = M^2 - M$ Suppose $a = M- x, b = M- y, c = M - z$ with $x,y,z$ small But now, see that $M^2 - M = a^2 + b^2 - c^2 - d^2$ $= (M-x)^2 + (M-y)^2 - (M-z)^2 - d^2$ $= M^2 - 2M(x+y-z) - d^2$ $\implies M(2x+2y-2z-1) = -d^2$, but since $M$ is much bigger, we must have both sides equal to zero. But then $2(x+y-z) = 1$, impossible. So, $q = 1$ is indeed impossible. So, the answer is $q \in \{-2,0,2\}$. $\blacksquare$

ISL marabot solve Claim: $q$ must be an integer. Proof: Suppose the napkin has the numbers $\{0,1,2\ldots,9\}$. We note that everything in the third line must be an integer, so $q\cdot 1\cdot 1=q$ must also be an integer. $\blacksquare$ Claim: $|q|\le 2$. Proof: Again suppose the napkin has the numbers $\{0,1,2,\ldots,9\}$. Then $81q$ is in the second line. Note that the largest number possible in the third line is $162$ and the smallest possible number in the third line is $-162$. Thus, $|q|\le 2$. $\blacksquare$ Claim: $q\ne -1$ and $q\ne1$ Proof: Suppose $q=-1$ or $q=1$. Then consider the ten real numbers $1,5,9,13,17,21,25,29,33,34$. We note that the first line can only have numbers $0,1,3,4,5,7\pmod 8$. Note that the two numbers $34-33=1$, $33-34=-1$, and $17-13=4$ are in the first line, so regardless of whether $q=1$ or $q=-1$, $4$ is in the second line. Now since $4$ must be in the third line, we need $a^2+b^2-c^2-d^2\equiv 4\pmod 8$, but there is a constraint: $a^2$, $b^2$, $c^2$, $d^2$ can only be $0$ or $1$ mod $8$. It's easy to see now that in mod $8$, $-2\le a^2+b^2-c^2-d^2\le 2$, and $4$ is not within that range. $\blacksquare$ Case 1: $\boxed{q=0}$. It's obvious only $0$ is written in the second line. Setting $a=b=c=d$ gives $0$ is also in the third line. Case 2: $\boxed{q=2}$. Let $a, b, c, d$ be four real numbers written on the napkin. We note \[(a-b)^2+(c-d)^2-(b-c)^2-(d-a)^2=2bc+2ad-2ab-2cd=2(b-d)(c-a)\] Now every number in the second line can be written as $2(b-d)(c-a)$ for some $a,b,c,d$, so everything in the second line is also in the third line. Case 3: $\boxed{q=-2}$. Define $a,b,c,d$ as the above case. Note\[(a-b)^2+(c-d)^2-(b-c)^2-(d-a)^2=2bc+2ad-2ab-2cd=2(b-d)(c-a)=-2(d-b)(c-a)\]Now every number in the second line can be written as $-2(d-b)(c-a)$ for some $a,b,c,d$, so everything in the second line is also in the third line.

The answer is $q=0,2,-2$. $q=0$ obviously works by taking $(a,b)=(c,d)$, and we have $(a-d)^2+(b-c)^2-(a-c)^2-(b-d)^2=2(a-b)(c-d)$, so $q=2,-2$ work. We now show that these are the only ones. Clearly $q$ works iff $-q$ works. Now, by writing $0,\ldots,9$ on the first line, every number on the third line is at most $2\cdot 9^2=162$, while $81q$ and $-81q$ are written on the second line, hence $|q|<2$. But since $q$ is also written on the second line and the third line is all integers, $q$ must be an integer, so $q \in \{-2,-1,0,1,2\}$. It now suffices to show that $q=1$ fails, since that means $q=-1$ fails as well. Let the numbers on the first line be $0,\ldots,7,\pi,e$, so $\pi e$ is on the second line. It is evident that the differences written down are $\pm(n),\pm(e-n),\pm(\pi-n),\pm(\pi-e)$ where $n$ must be an integer, so the only way we get a $\pi e$ term on the third line is from a term of the form $\pm(\pi-e)$. But in this case the coefficient of $\pi e$ is even, so the coefficient of $\pi e$ in $a^2+b^2-c^2-d^2$ is always even, and thus it can never equal $\pi e$. Thus $q=1$ fails, as desired. $\blacksquare$

We claim that $q \in \{0,2,-2\}$. $q=0$ trivially works, and $|q|=2$ works since $(a-d)^2+(b-c)^2-(a-c)^2-(b-d)^2=2(a-b)(c-d)$. Now, fix some real numbers $\{a_1<a_2<...<a_9\}$ chosen by Gugu. Pick an $x$ big enough such that $(x-a_9)^2>\frac{2(a_i-a_j)(a_k-a_l)}{q}$ for all $1 \leq i,j,k,l \leq 9$ and $(x-a_9)^2>\frac{2|(x-a_j)(a_k-a_l)|}{q}$ for all $1 \leq j,k,l \leq 9$. Now, observe that for all $a$ on the first line, $qa^2$ belongs to the second line, so $qa^2$ also belongs to the third line. From our choice of $x$, we have that $q(x-a_i)^2$ must be equal to $\pm 2|x-a_j||x-a_k|$. Since the expressions of $\pm 2|t-a_j||t-a_k|$ and $(t-a_i)^2$ are polynomials in $t$ and there are finitely many $a_i$, we can pick $x$ large enough such that $q(x-a_i)^2 \neq \pm 2|x-a_j||x-a_k|$ when $j,k \neq i$. Hence, we must have $q(x-a_i)^2= \pm 2(x-a_i)^2 \implies q = \pm 2$, as desired. $\blacksquare$

If the numbers on Gugu's napkin are $1,2,3,4,\dots,10$ then all of the numbers on the third line are integers, and $q\cdot 1\cdot 1$ must be from the third line, so $q$ is an integer. Let the largest number in the first line be $M$ then the largest number in the fourth line is $2M^2$. If $|q|>2$ then $q\cdot M\cdot (\pm M)$ will exceed $2M^2$. Thus, $q=-2,-1,0,1$ or $2$. Clearly, $q=0$ works, and $q=\pm 2$ works because \[(a-b)(c-d)=(a-d)^2+(b-c)^2-(a-c)^2-(b-d)^2.\]We claim that $q=\pm 1$ do not work. Clearly, if one of them works then the other must also work, so we only need to show that $q=1$ doesn't work. Let the numbers on the napkin be $0,1,2,3,4,\dots, 8,\pi$. Line $1$ has numbers in \[\{s\in \mathbb Z\mid -8\le s\le 8\}\cup \{\pm (\pi-s)\mid 0\le s\le 8\}.\]Note that $\pi$ is in the second line. We will show that $\pi$ is not in the third. For each number $a^2$, the $\pi$-term must be even, so no sum of numbers $a^2$ where $a$ is on line 1 can be $\pi$.

after getting $q$ integral it should also work to take $\{\sqrt{2},\sqrt{3},\dots,\sqrt{29}\}$ with the first ten primes assuming i'm not trolling; this immediately yields only $q\in \{-2,0,2\}$ as valid.

We should have a competition to see who can come up with the ugliest $q \in \{-1, 1\}$ counterexample. The answer is $q \in \{-2, 0, 2\}$. Clearly $q = 0$ works, and to see $q = \pm 2$ works, let the numbers be $a_1, \dots, a_{10}$, and consider any $q (a_i-a_j)(a_k-a_\ell)$ in the second row with $i, j, k, \ell \in \{1, 2, \dots, 10\}$. Note that \[(a_i-a_\ell)^2 + (a_j-a_k)^2-(a_i-a_)^2-(a_j-a_\ell)^2 = 2(a_ia_k-a_ja_k+a_ja_\ell-a_ia_\ell) = 2(a_i-a_j)(a_k-a_\ell).\]To get $q = -2$, simply reverse the signs. Now, by setting all $a_i$ to be integers, $q$ must be an integer. Furthermore, by considering the $a_i$ of the largest magnitude, it follows that $a^2+b^2-c^2-d^2 \leq 2a_i^2$ for $a, b, c, d \in \{a_1, a_2, \dots, a_{10}\}$, hence we have $|q| \leq 2$. It remains to check that we cannot have $q \in \{-1, 1\}$. It suffices to check that $q \neq 1$. We will offer an unbelievably lazy and ugly proof: just let $\{a_1, a_2, \dots, a_{10}\} = \left\{0, 1, 2, \dots, 8, 10^{100}\right\}$ and consider the number $10^{100} - 1$ in row $2$. Suppose for the sake of contradiction that there exists four differences $a, b, c, d$ such that $a^2+b^2-c^2-d^2 = 10^{100} - 1$. As $a \geq 8$ clearly, we must have $a = 10^{100} - x$ for some $x \in \{0, 1, \dots, 8\}$. If $c, d \leq 8$, then $a^2+b^2-c^2-d^2 > 10^{99}$, so we must have (WLOG) $c = 10^{100} - z$ for some $z \in \{0, 1, \dots, 8\}$. Observe that we cannot have only one of $b \geq 10^{100} - 8$ and $d \geq 10^{100} - 8$, so we split into cases based on whether the assertion is true for both or neither of $b$ and $d$. If $b \leq 8$ and $d \leq 8$, then \[a^2+b^2-c^2-d^2 = 2 \cdot 10^{100} (x-z) + x^2 + b^2 - z^2 - d^2.\]This is either in $[-128, 128]$ if $x = z$ and has absolute value greater than $2 \cdot 10^{100} - 128 > 10^{100} - 1$ if $x \neq z$, hence contradiction. Otherwise, let $b = 10^{100} - y$ and $d = 10^{100} - w$ for $y, w \in \{0, 1, \dots, 8\}$. We have \[a^2+b^2-c^2-d^2 = 10^{100} (x+y-z-w) + x^2+y^2-z^2-w^2\]which is again in $[-128, 128]$ if $x+y = z+w$ and has absolute value at least $2 \cdot 10^{100} - 128$ if $x+y \neq z+w$, contradiction. So we have shown a counterexample for $q = \pm 1$, proving that $q \in \{-2, 0, 2\}$ are the only possible values.

Ew Answer: $q=-2,0,2$. For the construction, $q=0$ is trivial, and for $q=2$ use $2(a-b)(c-d)=(a-d)^2+(b-c)^2-(a-c)^2-(b-d)^2$. Note that $q$ must be an integer; if not, make a number on the second line non integral (say using $q*1*1$ by placing $2$ numbers on the napkin that differ by $1$). Also, $|q|>2$ fails by considering $qa^2>2a^2$ where $a$ is the largest number on the first line. Lastly, to show $1$ and $-1$ fail, consider the napkin $1,2,10,20,30,40,50,60,70,80$, where $10$ is written on the second line. As $10\equiv 2 \pmod 4$, we have that $a,b$ are odd and $c,d$ even or vice versa. However, inspection gives $a^2 \equiv 1 \pmod 10$ if $a$ is odd. Similarly, $c^2 \equiv 0,4$ if $c$ is even. If $a,b$ are odd then $a^2+b^2-c^2-d^2\equiv 2-c^2-d^2 \equiv 2, 8, 4 \pmod {10}$, none of which are $0$, so $10$ can’t be on the third line.

Pretty easy First we prove that all $q\ne -2,0,2$ fail. Take numbers $\{\sqrt{2},\sqrt{3},\dots,\sqrt{29}\}$, which are the square roots of the first ten prime numbers. The number $q(\sqrt{2}-\sqrt{3})(\sqrt{5}-\sqrt{7})$ is on the second line, and must therefore be on the third line. Assume $q\ne 0$. In this case, due to square roots being linearly independent, it follows that $|a|$, $|b|$, $|c|$, and $|d|$ must be $\sqrt{5}-\sqrt{2}$, $\sqrt{7}-\sqrt{2}$, $\sqrt{5}-\sqrt{3}$, and $\sqrt{7}-\sqrt{3}$. In this case we easily find that $q=-2$ or $q=2$. Now we prove that $q=-2$, $q=2$, and $q=0$ work. If $q=0$ we are done; take $a=b=c=d$ on the third line. If $q=2$, write $2(a-b)(c-d)=(a-d)^2+(b-c)^2-(a-c)^2-(b-d)^2$, done. If $q=-2$, we have the same as above, except negating both sides of the equation, done. So we are done. $\blacksquare$ oh wait i actually didn't prove q was integral what a sell This only works if $q$ is integral, which is true: simply take $\{1,2,\dots,10\}$, and realize that $q$ is on the second line---and the third line is only integers.