Assume that $xy=a^2$ and $zt=b^2$ and that $z = max\{x,y,z,t\}$ Then $$xy-zt=(a-b)(a+b)=x+y=xy-z(x+y-z)=(z-y)(z-x)$$so $(a-b)(a+b)=(z-x)(z-y)$ From 4 numbers lemma we get $m,n,p,q \in \mathbb{N^\ast}$ such that $a-b = mn , a+b = pq , z-x = mp , z-y = nq$ Then $x+y= mnpq , 4a^2=4xy=(mn+pq)^2 $ and $y-x=mp-nq$ Since $4xy = (x+y)^2-(x-y)^2$ then $$(mn+pq)^2 = (mnpq)^2-(mp-nq)^2 \iff (mnpq)^2 = (m^2+q^2)(n^2+p^2)$$Claim $(m-1)(q-1) = 0$ and $(p-1)(n-1) = 0$ Proof If $m,n,p,q > 1 $ then $m^2+q^2<m^2q^2$ and $p^2q^2<p^2+q^2$ which is a contradiction. Then one of them is $1$ . Assume that $(p-1)(n-1) \neq 0 $ or $(m-1)(q-1) \neq 0$. WLOG $m = 1$ , $n>1 , p>1$ Then $(n^2-2)(p^2-2) \ge 4 \iff n^2p^2 \ge 2(n^2+p^2) \Rightarrow q^2+1 \ge 2q^2$ so we must have $q=1$ So $n^2p^2 = 2(n^2+p^2) \iff (n^2-2)(p^2-2) = 4 \Rightarrow p=q=2$ . Then $x+y = mnpq = 4$ and $xy = a^2 = 4$ so $zt = xy -x-y=4-4=0$ which is false. Hence our claim is proved. WLOG $m=1$ and $n=1$ Then $(pq)^2 = (p^2+1)(q^2+1) = p^2q^2 + p^2 + q^2 +1$ which can not hold. So $xy$ and $zt$ can not be both perfect squares.

Here is a number-theoretic solution The answer is no. Take $k_1=\gcd(x,y)$. Thus $x=k_1a^2$ and $y=k_1b^2$ for some $a,b$. Similarly, take $l_1=\gcd(z,t)$, $z=l_1c^2$ and $t=l_1d^2$. Let $g=\gcd(k_1,l_1)$, $k_1=gk$ and $l_1=gl$. The equation is transformed to $$(kab)^2-(lcd)^2\mid k(a^2+b^2) = l(c^2+d^2)$$where $\gcd(a,b)=\gcd(c,d)=\gcd(k,l)=1$. First, note that $k\mid l(c^2+d^2)$ thus $k\mid c^2+d^2$. Similarly $l\mid a^2+b^2$. As $\gcd(a,b)=\gcd(c,d)=1$, we conclude that each prime divisor of $k,l$ is either $2$ or $\equiv 1\pmod 4$. Now we do some parity check. By $\pmod 4$, $\nu_2(a^2+b^2) \leqslant 1$ and $\nu_2(c^2+d^2)\leqslant 1$. Thus by $\gcd(k,l)=1$, we also get $\nu_2(k)\leqslant 1$ and $\nu_2(l)\leqslant 1$. As $2$ cannot divide both $k,l$, we get $\nu_2(k(a^2+b^2))\leqslant 1$. Now, observe that as $(kab)^2-(lcd)^2$ must divide $k(a^2+b^2)$ and cannot be congruent to $2\pmod 4$, this number must be odd. This means all prime factor of $(kab)^2-(lcd)^2$ must be $\equiv 1\pmod 4$. Thus $$\begin{cases} kab+lcd &\equiv 1\pmod 4 \\ kab-lcd &\equiv 1\pmod 4 \\ \end{cases}\implies 2kab\equiv 2\pmod 4\implies k,a,b\text{ are odd.}$$Now we will further factorize $(kab)^2-(lcd)^2$. Before that, note that $\gcd(k,l)=1$ imply $k(a^2+b^2)=l(c^2+d^2)=T$ where $T=\mathrm{lcm}(a^2+b^2,c^2+d^2)$. Thus \begin{align*} \underbrace{\frac{T}{(kab)^2-(lcd)^2}}_{\in\mathbb{Z}} &= \frac{T}{T^2\left(\frac{ab}{a^2+b^2}+\frac{cd}{c^2+d^2}\right) \left(\frac{ab}{a^2+b^2}-\frac{cd}{c^2+d^2}\right)} \\ &= \frac{(a^2+b^2)^2(c^2+d^2)^2}{T(ac+bd)(ac-bd)(ad+bc)(ad-bc)} \end{align*}By permuting $a,b$ if necessary, WLOG $ac-bd>0$. Thus each prime factor of $ac+bd$, $ac-bd$, $ad+bc$, $ad-bc$ is $2$ or $\equiv 1\pmod 4$. Case 1: $c$ is odd, $d$ is even. This means each of $ac+bd$, $ac-bd$, $ad+bc$, $ad-bc$ are odd. Thus they are all $\equiv 1\pmod 4$. Hence $2ad\equiv 2\pmod 4$, contradiction as $d$ is even. Case 2: $c$ and $d$ are odd. This means $2$ divides each of $T$, $ac+bd$, $ac-bd$, $ad+bc$ and $ad-bc$. Hence the denominator is divisible by $2^5$. But the exponent of $2$ in the numerator is $4$, contradiction.

+) South Korean TST #5 solved by only using some division and comparison, but much messier than #3 post

The answer is no. We'll prove a stronger claim, that $xyzt$ cannot be a perfect square. Let $s=x+y=z+t=xy-zt$. WLOG $x\leq y$ and $z\leq t$. Then $$s=x(s-x)-z(s-z)=(x-z)(s-x-z)$$Since $(x-z)+(s-x-z)\equiv s\pmod 2$, it follows that $x-z,s-x-z$ are both even. Since $x+z\leq s$, it follows that $x-z=2a$ and $s-x-z=2b$ for some positive integers $a,b$. Thus $s=4ab$. We set a system of equations $$\begin{cases}x-z=2a \\x+z=4ab-2b\end{cases}$$so $$(x,y,z,t)=(2ab+a-b,2ab-a+b,2ab-a-b,2ab+a+b)$$Then $xyzt=\left[4a^2b^2-(a-b)^2\right]\left[4a^2b^2-(a+b)^2\right]=(4a^2b^2-a^2-b^2)^2-(2ab)^2$ Since $xyzt$ is a perfect square, it follows that $$(4a^2b^2-a^2-b^2)^2-(2ab)^2\leq (4a^2b^2-a^2-b^2-1)^2$$$$\iff 4a^2b^2-2a^2-2b^2-1\leq 0$$$$\iff (2a^2-1)(2b^2-1)\leq 2$$Thus $a=b=1$. But then $z=0$, contradiction.

Assume such $x,y,z,t$ exist then set $m^2=xy$ and $n^2=zt$. Solving the equations we get: $$m^2-n^2=x+\frac{m^2}{x} \Rightarrow x^2-(m^2-n^2)x+m^2=0$$ We know the discriminant of this quadratic must be a perfect square. Doing a similar thing for $z$ we get: $$(m^2-n^2)^2-4m^2 \quad \text{and} \quad (m^2-n^2)^2-4n^2 \quad \text{are perfect squares}$$This means they're product is a perfect square so: $$(m^2-n^2)^4-4(m^2+n^2)(m^2-n^2)^2+16m^2n^2=\left( (m^2-n^2)^2-2(m^2+n^2+1) \right)^2+4(1+2m^2+2n^2)$$By considering the minimum distance between adjacent squares this gives: $$4(1+2m^2+2n^2) \geq 2\left | (m^2-n^2)^2-2(m^2+n^2+1) \right |+1$$If $m=n+1$ we have: $$(2m-1)^2-4m^2=1-4m$$is a perfect square which is a contradiction as $m \geq 1$. Therefore as $m>n$ (as $x,y$ positive) $m \geq n+2$ so we have: $$ (m^2-n^2)^2-2(m^2+n^2+1) \geq (4m-4)^2-4m^2-2 \geq 12m^2-32m+14$$Also: $$LHS \leq 8m^2+4$$Combining these results: $$8m^2+4 \geq 24m^2-64m+29 \Rightarrow 0 \geq 16m^2-64m+25=(4m-8)^2-39 $$Hence $m \leq 2$ or $4m-8 \leq 6 \Rightarrow m \leq 3$. As we've already shown $m \geq n+2 \geq 3$ (as $n$ positive) the only case left to consider if $m=3,n=1$ but: $$n=1 \Rightarrow z=t=1 \Rightarrow z+t=2 \neq m^2-n^2=8$$giving a contradiction. So it is impossible for $xy$ and $zt$ to both be perfect squares.

Claim: All the solutions to $xy-zt=x+y=z+t$ are of the form \[\{\{x,y\},\{z,t\}\}=\{\{2cd-d+c,2cd-c+d\},\{2cd-c-d,2cd+c+d\}\}\]where $c,d$ are positive integers. Proof: Let $k=xy-zt=x+y=z+t$. Then, we have \[x(k-x)-t(k-t)=k,\]or \[(x-t)(k-x-t)=k.\]Without loss of generality, suppose $x>t$ (if $x=t$, then $xy=zt$ so $k=0$, which isn't possible). Let $a=x-t$ be a positive integer. We must now have that \[x+t=k-k/a,\]so $a\mid k$, so let $k=ab$, where $b$ is a positive integer. Thus, \begin{align*} x-t &= a \\ x+t &= ab-b, \end{align*}so \begin{align*} x &= \frac{ab+a-b}{2} \\ t &= \frac{ab-a-b}{2}. \end{align*}This implies that $ab\equiv a+b\pmod{2}$, or that $(a-1)(b-1)\equiv 1\pmod{2}$, so both $a$ and $b$ are even. Let $a=2c$ and $b=2d$ where $c$ and $d$ are positive integers. We have that \[x=2cd+c-d,\,\,\, t=2cd-c-d,\,\,\, k=4cd,\]so the solutions are of the desired form. $\blacksquare$ Suppose that $xy$ and $zt$ were squares, and WLOG take $c>d$. We see then that there exist positive integers $a$ and $b$ such that \begin{align*} a^2 &= (2cd+(c-d))(2cd-(c-d)) \\ b^2 &= (2cd+(c+d))(2cd-(c+d)), \end{align*}or \begin{align*} a^2+(c-d)^2 &= (2cd)^2 \\ b^2+(c+d)^2 &= (2cd)^2. \end{align*}Subtracting the equations gives \[(a-b)(a+b)=4cd.\]This implies that $a-b>0$, and also that $a-b$ is even. If we had $a-b\ge 4$, then $a+b\le cd$, so $a\le cd$. Thus, \[(2cd)^2-(c-d)^2\le(cd)^2,\]or $|c-d|\ge\sqrt{3}|cd|$. This is clearly impossible as $|c-d|=c-d\le c$ and $\sqrt{3}|cd|=\sqrt{3}cd>c$, so we must have $a-b=2$. Thus, $a+b=2cd$, so $a=cd+1$ and $b=cd-1$. Plugging this back in, we get that \[c^2+d^2+1=3c^2d^2,\]or \[(3c^2-1)(3d^2-1)=10.\]It is easy to check that this has no solutions, so there are $\boxed{\mathrm{no}}$ solutions.

RaduAndreiLecoiu wrote: Wlog $x\ge y$ and $z\ge t$ Since the numbers are positive $xy > zt$ and $t = x+y-z$. Then $$xy-zt = xy - z(x+y-z) = (z-x)(z-y) > 0$$If $xy = a^2$ and $zt=b^2$ , $a \neq b$ we get that $$(a-b)(a+b) = (z-x)(z-y) = x+y $$Now apply the four number theorem. Then $$a-b = mn , a+b = pq , z-x = mp , z-y = nq $$$m,n,p,q > 0 $ So $a=xy=\frac{mn+pq}{2}$ and $x+y = mnpq$ But $$2mnpq \ge mn+pq \iff (mn-1)(pq-1) + mnpq \ge 1$$which is true since $mn \ge 1$ and $pq \ge 1$ So $ x+y \ge xy \iff (x-1)(y-1) \le 1$, so one of them is 1 , but $x\ge y \Rightarrow y=1$ Since $x = \frac{mn+pq}{2} = mnpq - 1 = a^2 $ then $$mn+pq = 2mnpq - 2 \iff mnpq + (mn-1)(pq-1) = 3 $$and $mnpq = a^2 +1$ Since $a \neq 0$ and $ a \ge 2 \iff mnpq \ge 5$ we must have $mnpq = 2$ Then $(mn-1)(pq-1) = 1$ , so $mn\neq 1$ and $pq\neq 1$ $\Rightarrow mnpq \ge 4$ , impossible. So $xy$ and $zt$ can't be both perfect squares . maybe a little mistakes here? $a=xy=\frac{mn+pq}{2}$

Claim: There exist $a,b$ such that $$(z,x,y,t)=\left(\frac{ab-(a+b)}{2},\frac{ab-(a-b)}{2},\frac{ab+(a-b)}{2},\frac{ab+(a+b)}{2}\right)$$ Proof: WLOG $x\le y\le z\le t$, and define $y-x=a,z-y=b,t-x=a+b$. Then, note that for any quadruple $(n,n+a,n+b,n+a+b)$, we have $(n+a)(n+b)-n(n+a+b)=ab$, which is constant. Therefore, if we want the equality relation in the problem, we must have $2n+a+b=ab\implies n=\frac{ab-a-b}{2}$. As $n$ is unique, this means that for a pair of differences $(a,b)$ such that $2|a,b$, there exists exactly one tuple $(z,x,y,t)$ which satisfies the problem assertion. Namely, $(z,x,y,t)=\left(\frac{ab-(a+b)}{2},\frac{ab-(a-b)}{2},\frac{ab+(a-b)}{2},\frac{ab+(a+b)}{2}\right)$ Suppose that we can have $xy,zt$ both be squares. Then, we need pair $(a,b)$ such that $(ab)^2-(a+b)^2$, $(ab)^2-(a-b)^2$ are both perfect squares. Letting $k=\gcd(a,b)$, $a=Ak, b=Bk$, we need $$k^2(AB)^2-(A+B)^2, k^2(AB)^2-(A-B)^2$$to both be squares. First, note that if $2|AB$, then both of these numbers will be $-1\pmod 4$, so neither can be square. So, $A,B\equiv 1\pmod 2$. Now, considering odd $p$, where $p|A,p\not | B$, we have that $-(A+B)^2$ is a quadratic residue mod $p$. As $A+B\neq 0\pmod p$, we must have $-1$ is a quadratic residue. Hence, $p\equiv 1\pmod 4$. This argument can be repeated for all prime factors of $A,B$, so $A\equiv B\equiv 1\pmod 4$. Now, $v_2(A+B)=1$, $v_2(A-B)\ge 2$. So, the only way such that $k^2(AB)^2-(A+B)^2$ and $k^2(AB)^2-(A-B)^2$ can have the same $v_2$ is if $k$ is odd. To finish, note that $k^2(AB)^2-(A+B)^2\equiv 1-4\equiv 5\pmod 8$, which is a contradiction.

So, we have $xy-zt=x+y=z+t$. Case $1$: $x+y$ is even, $x+y=2a$ for some integer $a$. Let $x=a-b, y=a+b, z=a-c, t=a+c$, WLOG assume $b,c$ non-negative integers(that is $y>x, t>z$). Then, $a=\frac{c^2-b^2}{2}$ from the condition. Let $c=b+k$. So $a=(bk+\frac{k^2}{2})$. We have, if both $xy,zt$ are perfect squares, then $a^2-b^2$ and $a^2-(b+k)^2$ are squares. Thus, $(2bk+k^2)^2-(2b)^2$ is a perfect square, and $(2bk+k^2)^2-(2b+2k)^2$ is a perfect square. Let $(2bk+k^2)=p$. Also let $(2bk+k^2)-(2b)^2=(p-e)^2, (2bk+k^2)-(2b+2k)^2=(p-e-f)^2$. Obviously $f$ is even. Since $x,y,z,t$ are positive integers, so are $k,c,p,e,f$($b$ might be $0$ though). Note that $(p-e)^2-(p-e-f)^2=4p$. Hence if $2(p-e)-4>p$ then $f \geq 4$ will be a contradiction as difference of the squares is too large so in this case $f=2$. So we have $2(p-e)+2=2p$ or $e=1$. So we have $p^2-4b^2=(p-1)^2$ contradiction. So, now assume $f \geq 4$ as $f$ is even. Thus $2(p-e)-4 \leq p$. So, $e\geq \frac{p-4}{2}$ thus $4(2b)^2>3(2bk)^2$. This means that since $k$ is a positive integer, $k<2$ and so $k=1$. So, $p=2b+1, c=b+1$. Thus $p^2-(2b)^2=4b+1, p^2-(2c)^2=(2b+1)^2-(2b+2)^2=-3-4b$. The latter term is negative since $b$ is positive integer so it cannot be a square. But this contradicts both the facts that $zt$ is positive and that squares are positive numbers. So, both cannot be simultaneously squares in this case. Case $2$: $x+y$ is odd, $x+y=2a+1=f$. Consider $2x=f+b, 2y=f-b, 2z=f+c, 2t=f-c$ assuming WLOG that $b,c$ are positive($x>y,z>t$ that is). Thus $c^2-b^2=4f$. Let $b+2m=c$. Obviously $m$ is positive integer and $f=bm+m^2$. Now we have that if $xy, zt$ are squares then $(bm+m^2)-b^2=(f-r)^2$ and $(bm+m^2)-(b+2m)^2=(f-r-s)^2$ for some integers $r,s$. Note that $(f-r)^2-(f-r-s)^2=4f$ which is analogous to the equation we got earlier. Use the exact same logic as earlier to show that no solutions exist here too. Proved

If $2 \nmid x+y$, then $2 \mid xy$ and similarly, $2 \mid zt$ so $2 \mid xy-zt$, which is impossible since $xy-zt=x+y$. So $x+y$ is even. Let the naturals $a,b,c,d,u$ be such that $$ x+y=2u, xy=u^2-a^2=b^2, zt=u^2-c^2=d^2, |a^2-c^2|=2u$$We wish to show that they do not exist. Claim: $a \neq d$. Suppose otherwise, then, $|a^2-b^2|=|(a-b)(a+b)|=2u$, but we clearly need $u<a+b<u+u=2u$, so $1<|a-b|<2$ which is impossible. Proven. Lemma: If distinct naturals $a,b,c,d,n$ satisfy $a^2+b^2=c^2+d^2=n$, then there exists integers $p,q,r,s$ such that either $$a=pr-qs, b=ps+qr, c=ps-qr, d=pr+qs$$or $$a=pr-qs, b=ps+qr, d=ps-qr, c=pr+qs$$Proof: Let $n=p_1p_2 \dots p_k$ be prime factors(possibly repeated) of $n$. Note the following: 1) If $a^2+b^2=n=n_1n_2$, then $n_1$ and $n_2$ are also sum of squares. 2) If $a^2+b^2=n=n_1n_2$, there exist $e,f,g,h$ such that $a=eg-fh, b=eh-fg$ and ${n_1}^2=e^2+f^2$, ${n_2}^2=g^2+h^2$. 3) A prime can only be written as a sum of squares in at most 1 way. So, the conclusion immediately follows for $k \leq 2$. It suffices to check for $k=3$. (Note: $n=({x_1}^2+{y_1}^2)({x_2}^2+{y_2}^2)({x_3}^2+{y_3}^2)$, suffices to check any 'combi' with the 3 others, details omitted) After that, we can induct from $k$ to $k+1$: Since by removing $p_{k+1}$, we can use the result for $k$ to write it as $$ n=(pr-qs)^2+(ps+qr)^2=(ps-qr)^2+(pr+qs)^2$$We reduce it to the $k=3$ case. Lemma proven. Case 1: $$a=pr-qs, b=ps+qr, c=ps-qr, d=pr+qs$$ $$a^2-c^2=(a-c)\cdot (a+c)=(p-q)(r+s)\cdot (p+q)(r-s)=(p^2-q^2)(r^2-s^2)=S=2u=2\sqrt{u^2}=2\sqrt{(p^2+q^2)(r^2+s^2)}$$Let $S_1=p^2-q^2$. If $S_1\neq 0$ we have either $\sqrt{p^2+q^2}\leq |S_1| <\sqrt{2(p^2+q^2)}$, which happens when $||p|-|q||=1$ or $|S_1|\geq 2\sqrt{(p^2+q^2)}$ Since $u^2=(p^2+q^2)(r^2+s^2)$, the only possible case is when $|p^2-q^2|=\sqrt{(p^2+q^2)}$, and $r^2-s^2=2\sqrt{(r^2+s^2)}$(or the constant 2 goes to the first one, which is the same thing WLOG). Implying that $(p,q,r,s)$ is a permutation of $(1,0,2,0)$ which is clearly impossible, as it renders at least 2 of $a,b,c,d$ having the same absolute value. Case 2: $$a=pr-qs, b=ps+qr, d=ps-qr, c=pr+qs$$ $$a^2-c^2=(a-c)\cdot (a+c)=2pr \cdot 2qs=4pqrs=2u=2\sqrt{u^2}=2\sqrt{(p^2+q^2)(r^2+s^2)}$$ Thankfully this is much easier than case 1 and I now have more time to celebrate Christmas. By AM-GM, we get $\sqrt{p^2+q^2}\geq \sqrt{2|pq|}$ so we must have $|p|=|q|$ and $|r|=|s|$ which again gives solutions where 2 numbers have the same absolute value. Hence it is impossible.

The answer is no. Let $(xy,zt) = (a^2,b^2)$. If $2 \nmid a - b$, then $x+y$ is odd and $2 \mid xy$. Similarly, $2 \mid zt$, a contradiction. Thus, $2 \mid a-b$. Now note $x+y = z+t = a^2-b^2$ and note \begin{align*} x,y \text{ integer} &\iff (x+y)^2 - 4xy &= (a^2-b^2)^2 - 4a^2 \quad &\text{is square; and} \\ z,t \text{ integer} &\iff (z+t)^2 - 4zt &= (a^2-b^2)^2 - 4b^2 \quad &\text{is square}. \end{align*}Shorthanding $(a,b) = (m+n,m-n)$ (because we know $a,b$ are the same parity), we wish to find positive integers $m$, $n$ for which \[4m^2n^2 - (m+n)^2, \quad 4m^2n^2 - (m-n)^2 \quad \text{are both squares}.\]Thus, their product is a square. However, note upon expansion the inequality \[(4m^2n^2 - m^2 - n^2 - 1)^2 < (4m^2n^2 - (m+n)^2)(4m^2n^2 - (m-n)^2) < (4m^2n^2 - m^2 - n^2 )^2.\]Therefore, we have no solutions.

No. Suppose there does exist some satisfactory $(x,y,z,t)$. We will consider a solution $(x,y,z,t)$ and determine what conditions it satisfies by solving the diophantine $xy-zt=x+y=z+t$. WLOG assume $x \ge y$ and $z \ge t$. Claim: The expression $x+y=z+t$ is not odd. Solution: Suppose otherwise. Then, we clearly have that at least one of $x,y$ is even, so $xy$ is even. Similarly, $zt$ is even. This implies that $xy-zt$ is even, a contradiction. $\fbox{}$ Now, let $x+y=z+t=2d$. Let $x=d+k,y=d-k, z = d+\ell, t = d-\ell$ for integers $k,\ell$ with $k,\ell \ge 0$. We have \[2d=xy-zt=d^2-k^2 - (d^2-\ell^2) = \ell^2-k^2.\]Observe that $\ell$ and $k$ are the same parity. Let $a = \frac{k+\ell}{2}, b = \frac{\ell - k}{2}$. Note that both $a$ and $b$ are integers and $a \ge b \ge 0$. Clearly, we have $b \ne 0$ because we would otherwise get $0 < x+y=2d=0$. Thus, we have $a\ge b > 0$. Observe that \[2d = (a+b)^2 - (a-b)^2 = 4ab,\]so we have the parametrization \[(x,y,z,t) = (2ab+a-b,2ab-a+b,2ab+a+b,2ab-a-b).\]Note that $(a,b)=(1,1)$ does not work because it gives $t=0$; thus, we have $a \ge 2$. Now, we have \[xy=4a^2b^2-a^2+2ab-b^2, zt = 4a^2b^2-a^2-2ab-b^2.\]Let $xy=m^2,zt=n^2$. We have $m^2-n^2 = 4ab$. It is clear that $m$ and $n$ have the same parity, so we get that $4ab=m^2-n^2 \ge 4n+4$, implying $n \le ab-1$. Then, we get that \[(ab-1)^2 \ge n = 4a^2b^2-(a+b)^2.\]Equivalently, we have \[a^2+2ab+b^2 = (a+b)^2 \ge 4a^2b^2-(ab-1)^2 = 3a^2b^2 + 2ab - 1,\]so \[a^2+b^2 \ge 3a^2b^2-1 > 2a^2b^2 = a^2b^2+a^2b^2 \ge a^2+b^2.\]This is clearly absurd, so we are done.

Solution with anantmudgal09. Key Claim: We can explicitly parametrize $(x,y,z,t)$ given the equations. Proof: Let $xy-zt=x+y=z+t=a$. Then $y=a-x,t=a-z$. We get the key factorization: \[ a=x(a-x)-z(a-z)=(x-z)(-x-z+a)=(x-z)(y-z). \]Let $a=cd$ with $x-z=c,y-z=d$. WLOG $c\ge d$. Then $x=z+c,y=z+d$. Now $2z+c+d=cd$, so $z=\tfrac{cd-c-d}{2}$. Then $y=z+d=\tfrac{cd-c+d}{2}$, and $x=z+c=\tfrac{cd+c-d}{2}$, and $t=cd-z=\tfrac{cd+c+d}{2}$. So we have the explicit parametrization \[ (x,y,z,t) = \left(\frac{cd+c-d}{2}, \frac{cd-c+d}{2}, \frac{cd-c-d}{2}, \frac{cd+c+d}{2}\right). \]This proves the claim. $\square$ Now that we have a parametrization, we check for the squares condition. Assume $xy$ and $zt$ are squares. We have \begin{align*} 4xy &= (cd+c-d)(cd-c+d) = (cd)^2-(c-d)^2=u^2 \\ 4zt &= (cd-c-d)(cd+c+d) = (cd)^2-(c+d)^2=v^2, \end{align*}for some $u,v$. The rest of the solution is to use size to conclude that there are no solutions. The following claim will help with size calculations. Claim: $u-4\ge v$. Proof: Clearly $u>v$. We claim $u$ and $v$ are of the same parity. This is clear from the definitions of $u$ and $v$, since $c-d$ and $c+d$ are of the same parity. It suffices to show $u-2>v$. Indeed, \begin{align*} (u-2)^2-v^2 &= \big[[(cd)^2-(c-d)^2]- 4\sqrt{(cd)^2-(c-d)^2} + 4\big] - \big[ (cd)^2-(c+d)^2\big] \\ &= 4cd - 4 \sqrt{(cd)^2-(c-d)^2} + 4 > 0. \end{align*}This proves the claim. $\square$ Claim: Let $\epsilon=cd-\sqrt{(cd)^2-(c-d)^2}$. Then $\epsilon < c$. Proof: We have \begin{align*} &\qquad \epsilon = cd-\sqrt{(cd)^2-(c-d)^2} < c \iff (cd)^2-(c-d)^2 > (cd-c)^2 \\ &\iff -(c-d)^2 > -2c^2d + c^2 \iff (c-d)^2 < c^2(2d-1), \end{align*}which is clearly true. $\square$ Now, \begin{align*} (u-4)^2-v^2 &= \big[[(cd)^2-(c-d)^2] - 8\sqrt{(cd)^2-(c-d)^2} + 16 \big] - \big[(cd)^2-(c+d)^2\big] \\ &= 4cd - 8\sqrt{(cd)^2-(c-d)^2} + 16 = 4cd - 8(cd - \epsilon) + 16 \\ &= -4cd + 8\epsilon + 16 < -4cd+8c+16=16+4c(2-d). \end{align*}Unless $c(d-2)\le 4$, we get a contradiction. Clearly $c,d>1$; if $c=1$ or $d=1$ then $v^2<0$. We need to check $(c,d)=(c,2),(3,3),(4,3)$. It is not hard to check that none of these work, so we conclude that there are no solutions.

Wait can someone check my sol? I think this is absurdly short, so I'm worried that it's wrong.

We first check that $xy - zt = x + y = z + t$ are all even. Suppose otherwise, and that $x + y = z + t$ are odd. Then, one of $(x, y)$ is even, and one of $(z, t)$ is even. Hence $xy - zt$ is even, a contradiction. Write $xy - zt = x + y = z + t = k$ hence $(x, y) = (k - a, k+a)$ and $(z, t) = (k - b, k + b)$ for some positive integers $a, b$. Plugging in, we get $b^2 - a^2 = 2k$, and we want to determine if $k^2 - a^2$ and $k^2 - b^2$ can both be perfect squares. Clearly $a, b$ must be of the same parity, or else their difference can not be even. So let $b = m+n, a = m - n$ where $m$ is greater. Indeed, we get $2mn = k$ and we wish to determine if it is possible for $(2mn)^2 - (m-n)^2$ and $(2mn)^2 - (m+n)^2$ to both be perfect squares. Now we do something really dumb. Multiply the two of them and check that the product actually can't be a perfect square. The product, after extensive multiplying, is\[P = 16m^4n^4- 8(m^2 + n^2)m^2n^2 + m^4 - 2m^2n^2 + n^4.\]Clearly this expression is strictly less than $(4m^2n^2 - m^2 - n^2)^2$ since $m, n$ must be positive. Then, the stupid part is that somehow\[(4m^2n^2 - m^2 - n^2 - 1)^2 < P\]holds since it simplifies to $4m^2n^2 - 2m^2 - 2n^2 - 1 > 0 \iff (2m^2 - 1)(2n^2 - 1) > 2$ for all $(m, n) \neq (1, 1)$. A quick plug in shows that $(m, n) = (1, 1)$ does not work. For all other cases, $P$ is bounded between two perfect squares, and hence cannot be a perfect square. So our answer is $\boxed{no}$. Remark: "Something really dumb" took like half a day to come with lol.

Assume that $xy=u^2, zt=v^2.$ We can assume $x\geq y, t\geq z$ and since $xy-zt>0,$ it follows that $x>z.$ We have also $$0<xy-zt=xy-z(x+y-z)=(x-z)(y-z),$$so in fact $t>x\geq y>z.$ Now write $x+y=xy-zt=xy-zx-zy+z^2$ as $$y(x-z-1)=x+zx-z^2.$$From here we obtain (given that $x-z-1>0$) $$x-z-1\mid x+zx-z^2+(-zx+z^2+z)+(-x+z+1)=2z+1,$$thus $x\leq 3z+2$ (note that this is also true when $x-z-1=0$)$.$ Next we see that $2\mid u-v.$ Indeed$,$ if this is not the case$,$ then WLOG $v$ is odd (the other case is the same)$,$ thus $z, t$ are odd and so $x+y$ is even$.$ We have that $xy$ is also even$,$ therefore $x, y$ are even$,$ but in the end it doesn't even matter ($2\nmid xy-zt=x+y$)$.$ Then there is the inequality $u=\sqrt{xy}>v=\sqrt{zt}>z.$ We surely cannot have $u=v, $ and by the previous paragraph it follows that $u\geq v+2.$ Now use $zt=v^2$ $$\Rightarrow v^2+z^2=z(x+y)=z(u^2-v^2)\geq z(4v+4),$$so $v^2-4zv+z^2-4z\geq 0.$ This is our favourite high-school quadratic inequality in $v,$ so our first job is to calculate the roots of the equation $v^2-4zv+z^2-4z=0,$ namely $$v_{1, 2}=\frac{4z\pm 2\sqrt{3z^2+4z}}{2}=2z \pm \sqrt{3z^2+4z}.$$We easily check that $2z-\sqrt{3z^2+4z}<z,$ so it must be the case $v\geq 2z+\sqrt{3z^2+4z}>3z.$ We are ready for the contradiction$.$ The final inequality is$:$ $$6z+4\geq 2x\geq x+y=u^2-v^2\geq 4+4v>4+12z,$$as desired$.$ $\blacksquare$

I have different solution. Can someone check this? Lemma-1. If $n\equiv 3\pmod 4$ then exist prime number $p\equiv 3\pmod 4$ that divides $n$. Lemma-2. if $p$ is prime number and divides $A^2 +1$ for some integer $A$ then $p=2$ or $p\equiv 1\pmod 4$ Suppose that $xy=B^2 , zt=A^2$ for some integer $A,B$. Case-1.$x,y,z,t$ are all odd. In this case we have $x+y=xy-zt\equiv {1-1}\equiv 0\pmod 4$ but $x+y\equiv {3+3}\equiv {1+1=2}\pmod 4$. Contradiction! Case-2. One of this is even. Suppose that $x$ is even. We have $xy-x-y-1=(x-1)(y-1)=zx+1=A^2 +1$ now if 4 divides $x$ then with Lemma-1 there exist prime number $p\equiv 3\pmod 4$ that divides $x-1$ and so divides $A^2 +1$ but with Lemma-2 this is Contradiction! It means $x\equiv 2\pmod 4$ but we know $xy$ is perfect square and it means $y$ is even number so we have $y\equiv 2\pmod 4$ too. $x+y=xy-zt$ and $x+y , xy$ are even numbers so $zt$ is even and with above proof we conclude $z\equiv t\equiv x\equiv y\equiv 2\pmod 4$ but in this case we have $xy-zt\equiv 0\equiv {x+y} \equiv 4 \pmod 8$ Contradiction! We are done.

Let $(xy,zt) = (a^2,b^2)$. If $2 \nmid a - b$, then $x+y$ is odd and $2 \mid xy$. Similarly, $2 \mid zt$, a contradiction. Thus, $2 \mid a-b$.Now note $x+y = z+t = a^2-b^2$ and note $x,y \text{ integer}\implies (x+y)^2 - 4xy = (a^2-b^2)^2 - 4a^2 \quad \text{is square}$ $z,t \text{ integer} \implies (z+t)^2 - 4zt = (a^2-b^2)^2 - 4b^2 \quad \text{is square}.$ $(a,b) = (m+n,m-n)$ (because we know $a,b$ are the same parity), we require to find to find positive integers $m$, $n$ for which$4m^2n^2 - (m+n)^2, \quad 4m^2n^2 - (m-n)^2$$ \quad \text{are both squares}.$ Thus, their product is a square. However, note upon expansion the inequality $(4m^2n^2 - m^2 - n^2 - 1)^2 $< $(4m^2n^2 - (m+n)^2)(4m^2n^2 - (m-n)^2)$$ < (4m^2n^2 - m^2 - n^2 )^2.$ Therefore, we have no solutions.

No. Assume otherwise. First we claim that the common value must be even. Assume not. Then $\{x,y\}$ and $\{z,t\}$ are distinct modulo $2$, so $xy$ and $zt$ are both even and so is their difference which gives a contradiction. $\square$ Let $x+y=2s=z+t$. Let $xy=a^2$ and $zt=b^2$. Define $c,d$ such that : $$ x=s+c \quad y=s-c \quad z=s+d \quad t = s -d $$ Note that the given conditions rearrange to : $$ s^2 = a^2+c^2=b^2 +d^2$$$$2s = a^2-b^2=d^2-c^2$$ The second equation tells us that $\{a,b\}$ and $\{c,d\}$ are of the same parity, so their differences must be atleast $2$. This gives $a+b \leq s$ and $c+d \leq s$. Squaring both sides gives $a^2+b^2 \leq s^2-2ab <s^2$. Similarly $c^2+d^2<s^2$. Now we have : $$2s^2=a^2+b^2+c^2+d^2 <s^2+s^2=2s^2$$ Contradiction. $\square$

r_ef wrote: I have different solution. Can someone check this? Lemma-1. If $n\equiv 3\pmod 4$ then exist prime number $p\equiv 3\pmod 4$ that divides $n$. Lemma-2. if $p$ is prime number and divides $A^2 +1$ for some integer $A$ then $p=2$ or $p\equiv 1\pmod 4$ Suppose that $xy=B^2 , zt=A^2$ for some integer $A,B$. Case-1.$x,y,z,t$ are all odd. In this case we have $x+y=xy-zt\equiv {1-1}\equiv 0\pmod 4$ but $x+y\equiv {3+3}\equiv {1+1=2}\pmod 4$. Contradiction! Case-2. One of this is even. Suppose that $x$ is even. We have $xy-x-y-1=(x-1)(y-1)=zx+1=A^2 +1$ now if 4 divides $x$ then with Lemma-1 there exist prime number $p\equiv 3\pmod 4$ that divides $x-1$ and so divides $A^2 +1$ but with Lemma-2 this is Contradiction! It means $x\equiv 2\pmod 4$ but we know $xy$ is perfect square and it means $y$ is even number so we have $y\equiv 2\pmod 4$ too. $x+y=xy-zt$ and $x+y , xy$ are even numbers so $zt$ is even and with above proof we conclude $z\equiv t\equiv x\equiv y\equiv 2\pmod 4$ but in this case we have $xy-zt\equiv 0\equiv {x+y} \equiv 4 \pmod 8$ Contradiction! We are done. I think this solution is correct but the step $xy-zt\equiv 0\equiv {x+y} \equiv 4 \pmod 8$ requires more explanation.

The answer is no, we now show it. CLAIM. $xy\equiv zt\pmod 2$ Proof. Otherwise, $x+y\equiv z+t\pmod 2$, hence $xy\equiv zt\equiv 1\pmod 2$, contradiction. $\blacksquare$ WLOG assume $z<t$ and $x>y$, then $x>z$. Suppose on the contrary that $xy=c^2,zt=d^2$, then $d\leq c-2$. Fixing $x,z$, we have $$t=x+\frac{x+z}{x-z-1},\hspace{20pt}y=z+\frac{x+z}{x-z-1}$$Let $\displaystyle k=\frac{x+z}{x-z-1}$, then $k(x-z-1)=x+z$, hence $$z=\frac{(k-1)x-k}{k+1}$$Therefore, $0\equiv (k-1)x-k\equiv 2x-1\pmod{k+1}$, hence $k$ is even, let $k=2m$, then $x\equiv m+1\pmod{2m+1}$. Let $x=a(2m+1)+m+1$, then $$z=\frac{(a(2m+1)+m+1)(2m-1)-2m}{2m+1}=a(2m-1)+m-1$$Therefore, we conclude that \begin{align*} zt&=(a(2m+1)+3m+1)(a(2m-1)+m-1)\\ xy&=(a(2m+1)+m+1)(a(2m-1)+3m-1)\\ \end{align*}Now it is just some simple bounding, notice that $m>0$, hence $a\geq 0$. Notice that $$4am+4m=xy-zt=c^2-d^2\geq c^2-(c-2)^2=4\sqrt{xy}-4$$If $a=0$ then we need $m\geq\sqrt{(3m+1)(m-1)}$, no solution. If $a=1$, we need $m\geq \sqrt{(5m+2)(3m-2)}$, no solution. If $a\geq 2$ then $\sqrt{xy}\geq a\sqrt{4m^2-1}\geq\sqrt{3}am$ Therefore, $$m+1\geq(\sqrt{3}-1)am\geq2(\sqrt{3}-1)m$$hence $m\leq 2$, which gives no solution.

The answer is NO. First let $s=x+y=z+t$. And we shall express $t=x+y-z$ and $x=z+t-y$. Giving us $2$ quadratic equations: $$z^2-z(x+y)+xy-x-y=0$$$$-y^2+y(z+t)-zt-z-t=0$$Let $ \Delta_1$ and $\Delta_2$ be the discriminants of the given quadratic equations, respectively. Thus we have that: $$\Delta_1^2=(x+y)^2-4xy+4x+4y=(x-y)^2+4(x+y)$$$$\Delta_2^2=(z+t)^2-4zt-4z-4t=(z-t)^2-4(z+t)$$Since we have that $x+y=z+t=s$, we have that $4s=\Delta_1^2-(x-y)^2=(z-t)^2-\Delta_2^2$. Thus we have that $\Delta_1^2+\Delta_2^2=(x-y)^2+(z-t)^2$. Now we state the obvious lemma. Lemma: If we have that $x^2+y^2=z^2+t^2$, then there exist $a,b,c,d$ such that $(x,y,z,w) \in \left\{(ac+bd,bc-ac,bc+ad,ac-bd)\right\}$ and all of their respected permutations. Now when we return to our problem and start input what we have, we easily come to the conclusion that $x-y=bc-ad$ and $z-t=ac+bd$, $\Delta_1 = bc+ad$ and $\Delta_2 = ac-bd$. When we now plug that in we get that $s=abcd$, and having that $s=x+y=z+t$, we get that: $$2x=bc-ad+abcd$$$$2y=abcd+ad-bc$$$$2z=ac+bd+abcd$$$$2t=abcd-ac-bd$$ From here on out we claim that $xy$ and $zt$ are both square numbers. If they are both square number then we must have that $4xy=m^2$ and $4zt=n^2$ are also square numbers. Plugging in what we got we have that: $$4xy-4zt=4s \implies (c^2-d^2)(a^2-b^2)=0$$thus we can claim wlog that $a=b$ Thus we have that $m^2=a^2(a^2c^2d^2-(d-c)^2)$ and $n^2=a^2(a^2c^2d^2-(d+c)^2)$ Thus we have that $a^2c^2d^2=(d-c)^2+k^2=(d+c)^2+g^2$. Applying our lemma again on the pairs $(d-c,k,d+c,g)$, we get a contradiction, since any solution we got, brought us that $xy$ and $zt$ aren't squares at the same time. Thus the answer is negative.

The answer is no. Done using the walk-through from NICE Journal Let $xy=a^2,zt=b^2$, then $x+y=a^2-b^2=z+t$. Note that $a^2-b^2 >0$, so $a\neq b$ \[(x-y)^2=(x+y)^2-4xy = (a^2-b^2)^2-4a^2\]\[(z-t)^2=(z+t)^2-4zt = (a^2-b^2)^2-4b^2\]Note that $xy,zt$ must be of the same parity mod 2. AFTSOC otherwise, then $xy-zt=x+y=z+t$ is odd, so both $xy,zt$ must be even, contradiction. Thus, we may write $a=m+n,b=m-n$, where $m>n>0$. Then, we now know that $(x-y)^2(z-t)^2$ is a square and equals: \[((a^2-b^2)^2-4a^2)((a^2-b^2)^2-4b^2)=(16m^2n^2-4(m+n)^2)(16m^2n^2-4(m+n)^2)\]Which equals, \[=16(4m^2n^2-m^2-n^2-2mn)(4m^2n^2-m^2-n^2-2mn)\]We can pull out $4^2$, so, \[S= (4m^2n^2-m^2-n^2)^2-(2mn)^2 \text{ is a square}\]Note that for $m>n>0$. \[ 0<(2m^2-1)(2n^2-1)-2 \Longrightarrow 4m^2n^2 < 8m^2n^2-2m^2-2n^2-1\]Thus, $S$ can be bounded between two consecutive squares \[(4m^2n^2-m^2-n^2-1)^2< (4m^2n^2-m^2-n^2)^2-(2mn)^2 < (4m^2n^2-m^2-n^2)^2\]thus, we have a contradiction, so $xy,zt$ cannot ever both be squares.$\blacksquare$.

Check $xy - zt = x + y = z + t$ are all even. Let $xy - zt = x + y = z + t = 2N.$ Take $x = N+a, y = N-a, z = N+b, t = N-b.$ We are given $b^2-a^2 = 2N.$ We have $b$ and $a$ are of equal parity. Suppose $N^2-a^2$ and $N^2-b^2$ are both perfect squares. Then take $b = g+h, a = g-h,$ so then $N = 2gh.$ But $$(N^2-a^2)(N^2-b^2)= ((2gh)^2-(g-h)^2)((2gh)^2-(g+h)^2) =(4g^2h^2-g^2-h^2)^2 - 4g^2h^2 $$is not a perfect square because it can be bounded between $(4g^2h^2-g^2-h^2-1)^2$ and $(4g^2h^2-g^2-h^2)^2.$ $\square$

The answer is no. Suppose otherwise; replace $x,y,z,t$ with $a,b,c,d$ in that order, and let $ab-cd=a+b=c+d=N$, where WLOG $a,c\leq N/2$ so $a>c$ (else $ab-cd<0$). Then we have $b=N-a$ and $d=N-c$, so $$a(N-a)-c(N-c)=N \implies N=\frac{a^2-c^2}{a-c-1} \implies N-a-c=\frac{a+c}{a-c-1},$$so we want $a(N-a)=ac+a\cdot \tfrac{a+c}{a-c-1}$ and $c(N-c)=ac+c\cdot \tfrac{a+c}{a-c-1}$ to be perfect squares. Evidently, $a$ and $c$ are the same parity, otherwise $a-c-1$ is even but $a+c$ is odd so we have $a-c-1 \nmid a+c$. As such, we can let $a+c=2x,a-c=2y$, so $x>y>0$ and we have $a=x+y,c=x-y$. Then $$\frac{a+c}{a-c-1}=\frac{2x}{2y-1}:=2k~(k \in \mathbb{Z}),$$so $x=k(2y-1)>0$. Then, $$ac+a\cdot \frac{a+c}{a-c-1}=x^2-y^2+(x+y)\frac{2x}{2y-1}=4k^2y^2-(k-y)^2:=(2ky-A)^2,$$and $$ac+c\cdot \frac{a+c}{a-c-1}=x^2-y^2+(x-y)\frac{2x}{2y-1}=4k^2y^2-(k+y)^2:=(2ky-B)^2,$$where $A<B<2ky$ are nonnegative integers. Then \begin{align*} (2ky-A)^2-(2ky-B)^2&=(k+y)^2-(k-y)^2\\ (B-A)4ky-(B^2-A^2)&=4ky\\ A+B&=\frac{B-A-1}{B-A}4ky\geq 2ky, \end{align*}where the last line comes from the fact that $B-A=1$ makes the LHS zero, while $A=B=0$ clearly fails as $k+y>0 \implies B \geq 1$. This means that $B \geq ky$, so $$4k^2y^2-(k+y)^2 \leq k^2y^2 \iff (k+y)^2 \geq 3k^2y^2.$$For a fixed value of $k+y$, the RHS is minimized when one of $k,y$ is equal to $1$, so setting $k=1$, we need $y^2+2y+1\geq 3y^2$, which only holds for $y=1$. Then $x=1(2-1)=1$, but this implies that $c=x-y=0$—impossible. As such, no such positive integers $(a,b,c,d)$ exist. $\blacksquare$

Author of this problem - my friend Mikhail Murashkin.

Under conditions of problem we have: $$xy-zt=x+y=z+t=(x-z)(y-z)=(x-t)(y-t).$$After division on all possible common divisors we can go to the problem: $$xy-zt=(z-x)(z-y)=(t-x)(t-y)|x+y=z+t$$for positive integers $x,y,xz,t$ with $(x,z)=(x,t)=(y,z)=(y,t)=1.$ Let $z=ak^2, t=al^2, x=cm^2, y=cn^2$, where $(a,c)=(k,l)=(m,n)=1$. Let's rewrite previuos chain as: $$(cmn-akl)(cmn+akl)=(ak^2-cm^2)(ak^2-cn^2)=(al^2-cn^2)(al^2-cn^2)|a(k^2+l^2)=c(m^2+n^2)$$. So it is true that: $$t=(cmn-akl)(cmn+akl)=$$$$=(ak^2-cm^2)(ak^2-cn^2)=(al^2-cn^2)(al^2-cn^2), \eqno(1)$$$$s=a(k^2+l^2)=c(m^2+n^2), \eqno(2)$$$$t|s. \eqno(3)$$For every prime $p$, dividing $t$, it is also true that$p|s$ from $(3)$ and $p\not | a,c,k,l,m,n$ from $(1),(2)$. Particularly, $2\not |t$, otherwise $a,c,k,l,m,n$ would be odd, so $4$ should divide $t$ but not $s$ - contradiction with $(3)$. It follows that $$(cmn-akl,cmn+akl)=(ak^2-cm^2,ak^2-cn^2)=1,$$and there are coprime positive integers $T_1,T_2,T_3,T_4$ such that $$cmn-akl = T_1T_2, \eqno(4)$$$$cmn+akl = T_3T_4, \eqno(5)$$$$ak^2-cm^2=cn^2-al^2=T_1T_3, \eqno(6)$$$$ak^2-cn^2=cm^2-al^2=T_2T_4. \eqno(7)$$ Further, $$(cmn-akl)^2-ac(kn-lm)^2=(ak^2-cm^2)(al^2-cn^2)=-T_1^2T_3^2,$$therefore it follows from $(4),(6)$ that $$ac(kn-lm)^2=T_1^2(T_2^2+T_3^2). \eqno(8)$$Analogously, $$(cmn+akl)^2-ac(km+ln)^2=(ak^2-cn^2)(al^2-cm^2)=-T_2^2T_4^2,$$and we get from $(5),(7)$ that $$ac(km+ln)^2=T_4^2(T_2^2+T_3^2). \eqno(9)$$Adding $(8)$ to $(9)$, we find that $$(T_1^2+T_4^2)(T_2^2+T_3^2)=ac((kn-lm)^2+(km+ln)^2)=a(k^2+l^2)c(n^2+n^2)=s^2,$$therefore the next divisibility is true from $(3)$ $$t^2=T_1^2T_2^2T_3^2T_4^2|(T_1^2+T_4^2)(T_2^2+T_3^2)=s^2.$$Let us remember that $t$ is odd, $s$ then should be even, so divisibility below is also true $$4T_1^2T_2^2T_3^2T_4^2|(T_1^2+T_4^2)(T_2^2+T_3^2). \eqno(10)$$It is easy to see that $(10)$ is true iff$T_1=T_2=T_3=T_4=1$, that contradicts to $(4),(5)$, q.e.d.

No. Suppose otherwise. Let $x=t+a$, $y=z-a$ for some integer $a$. Then $(z-a)(t+a)-zt=a(z-t-a)=z+t$. Let $b=z-t-a$, then $z+t=ab$ and $z-t=a+b$. It follows that $(ab)^2-(a+b)^2=4zt$, which is a perfect square. Similarly, $y+x=ab$, \[y-x=z-a-(t+a)=z-t-2a=b-a \implies (ab)^2-(a-b)^2=4xy\]another perfect square. Next, let $4zt=(ab-k)^2$ and $4xy=(ab-k+l)^2$. Then \[(ab-k+l)^2-(ab-k)^2=4ab \implies 2l(ab-k)+l^2 = 4ab\] Claim: $k \le ab/2$ Proof: Suppose otherwise. Then \[(ab)^2-(a+b)^2 < (\frac{1}{2}ab)^2 \implies \frac{3}{4}ab^2 < (a+b)^2 \implies a+b > \frac{\sqrt{3}}{2}ab\]Note that $ab > a+b$, else $xy$ would be nonpositive. In particular, at least one of $a,b$ must be at least $3$, and both must be at least $2$. Thus, WLOG let $a \ge 3$. But dividing by $a$ we have \[1+\frac{b}{a} > \frac{\sqrt{3}}{2}b \implies 1+\frac{b}{3} > \frac{\sqrt{3}}{2}b \implies \left( \frac{\sqrt{3}}{2}-\frac{1}{3} \right) b -1 < 0 \]But after rearranging it is apparent that this implies $b < 2$, contradiction. Returning to the problem, we see that $2l(ab-k)+l^2 = 4ab \implies 4ab > 2l(ab-k) > lab \implies l<4$. Furthermore, $l$ is even for parity reasons. Thus $l=2$. Hence $4ab-4k+4=4ab \implies k=1$. But then $(ab)^2-(a-b)^2=(ab+1)^2$,contradiction and we are done.

I think I have a different solution, does this work? WLOG assume $x\ge y.$ Then $x$ cannot be smaller than both $z$ and $t$, as this would mean $xy-zt\le 0,$ impossible. So assume $x\ge z$ as well. We have $$x+y=xy-zt=xy-z(x+y-z)=xy-zx-yz+z^2=(x-z)(y-z).$$If $y-z=1,$ then $x-z=x+y,$ absurd. Thus $x-y<x-z\le\tfrac{x+y}{2},$ so $x<3y.$ This gives the bound $$\frac{x+y}{\sqrt{xy}}=\frac{y(\tfrac{x}{y}+1)}{y\sqrt{\tfrac{x}{y}}}\le\frac{3+1}{\sqrt{3}}=\frac{4}{\sqrt{3}}.$$Now assume $xy$ and $zt$ are both perfect squares. Then there exists $m\le\sqrt{xy}$ such that $$xy-x-y=zt=(\sqrt{xy}-m)^2=xy-2m\sqrt{xy}+m^2.$$Rearranging, $$m\le 2m-\frac{m^2}{\sqrt{xy}}=\frac{x+y}{\sqrt{xy}}<\frac{4}{\sqrt{3}}.$$Hence $m=1$ or $m=2,$ so we have reduced the problem to a finite case check.

Let $n = xy - zt = x + y = z + t$. Then we can subsitute to get $n = x(n - x) - z(n - z) = n(x - z) + z^2 - x^2$ such $n \ge x, z$. As such, $n = n(x - z) + z^2 - x^2 = (n - x - z)(x - z)$ and $z^2 \equiv x^2 \pmod{n}$. Then, we need $n$ to be even, so $x - z, x + z$ must both be even, and thus $4 \mid n$. Then, we can take $n$ to be a multiple of $4$, then let $4ab = n$, so \begin{align*} z &= 2ab - a - b \\ t &= 2ab + a + b \\ x &= 2ab - a + b \\ y &= 2ab + a - b \end{align*}Then, $xy = 4a^2b^2 - (a - b)^2$ and $zt = 4a^2b^2 - (a + b)^2$ are perfect squares, and their difference $xy - zt = 4ab$. As such, by difference of squares, $\sqrt{xy} + \sqrt{zt} \le 2ab$ and $xy + zt \le 4a^2b^2$. Similarily, $(a - b)^2 + (a + b)^2 = 2a^2 + 2b^2 < 4a^2b^2$ holds when not both $a, b$ are $1$. However, $xy + zt + (a - b)^2 + (a + b)^2 = 8a^2b^2$, contradiction.

Very Nice problem! We claim that the answer is no and the product $xyzt$ cannot be a perfect square. Let $p = xy - zt = x + y = z + t$; WLOG $x \leq y,z \leq t$ It is easy to see that $4 | p$ by parity arguements, let $$p=xy-zt=x(p-x)-z(p-z)=px-pz-x^2+z^2 = (p-x-z)(x-z)$$,notice that $2$ must divide both of these. Let $(x-z)=2m$, $p-x-z = 2n$. It follows that $p=4mn$ and $(x,y,z,t)=(2mn+m+n,2mn-m-n,2mn+m-n,2mn+n-m)$ Now $xyzt = (2mn+m+n)(2mn-m-n)(2mn-m+n)(2mn+m-n)$ $=(4m^2n^2-(m+n)^2)(4m^2n^2-(m-n)^2)=(4m^2n^2-m^2-n^2)^2-(2mn)^2$ now to show that this is not a perfect square. We use bounding, $(4m^2n^2-m^2-n^2)^2-(2mn)^2 \leq (4m^2n^2-m^2-n^2-1)^2$ $\leftrightarrow (4m^2n^2-m^2-n^2)^2-(2mn)^2 \leq (4m^2n^2-m^2-n^2)^2-2(4m^2n^2-m^2-n^2)+1$ $\leftrightarrow (2m^2-2)(2n^2-2) \leq 2$ which is a contradiction!. EDIT : 150th HSO post!

It is not possible. Notice $4(xy-zt)-2(x+y)-2(z+t)=(2x-1)(2y-1)-(2z+1)(2t+1)=0$ so there exist positive integers $p,q,r,s$ such that $pq=2x-1,rs=2y-1,pr=2z+1,qs=2t+1.$ Then from $x+y=z+t$ we get $pq+rs+4=pr+qs,$ which rearranges to $(p-s)(r-q)=4.$ Then notice that $p,q,r,s$ are all odd, so $p-s,r-q$ are even so by symmetry we can assume they are both $2,$ as opposed to $-2.$ Then letting $p=2a+1,s=2a-1,r=2b+1,q=2b-1$ we get $x=2ab-a+b,y=2ab+a-b,z=2ab+a+b,t=2ab-a-b.$ Then we have $xy=4a^2b^2-(a-b)^2$ and $zt=4a^2b^2-(a+b)^2.$ First, we notice that $a=b=1$ is invalid and for all other pairs $a,b$ of positive integers we have $ab\sqrt3>a+b,$ so $4a^2b^2-(a+b)^2>a^2b^2.$ Thus if we let $xy=i^2$ and $zt=j^2$ for positive integers $i,j$ we have that $a^2b^2<zt<xy \le 4a^2b^2$ and $xy-zt=4ab.$ Thus if $i-j=1$ we see that $xy-zt=i+j$ but $j<i\le 2ab$ so $i+j<4ab.$ Now if $i-j \ge 2$ we see that $i>j>ab$ so $i+j>2ab$ so $xy-zt>4ab$ so we are done.

It's obvious, that $x+y = 2m$. So let $x = m-k, y = m+k, z = m-l, t = m+l, xy = a^2, zt = b^2$. We know that $2m = a^2 - b^2 = (m-k)(m+k) - (m-l)(m+l) = l^2 - k^2$ and $a^2 + k^2 = m^2 = b^2 + l^2$. In this format $(a, b, k, l)$ is equal to $(l, k, b, a)$, so we can assume that $l - k \leqslant a-b \implies l+k \geqslant a+b$. Let's denote $d_2 = l-k, d_1 = l+k$. Easy to see that $d_1 - d_2 \geqslant 2b$ and $2|d_1, d_2$. So, $4d_1^2 \leqslant (d_1d_2)^2 = (2b)^2 + (2l)^2 \leqslant (d_1 - d_2)^2 + (d_1+d_2)^2 = 2(d_1^2 + d_2^2) \leqslant 4d_1^2$. This can only happen if $d_1 = d_2 = 2$, but $4 = 2m > 2l = d_1 + d_2 = 4$, contradiction.

Let $xy=a^2$ and $zt=b^2$. Clearly, $xy>zt$ but $x+y=z+t$ which means that $\max{z,t}>\max{x,y}$. WLOG, assume $z>t$. Then, \[(a-b)(a+b)=a^2-b^2=xy-zt=xy-z(x+y-z)=(z-x)(z-y)\]so by the four-number theorem we can write $z-x=pq$, $z-y=rs$, $a-b=pr$, and $a+b=qs$ for positive integers $p,q,r,s$. Then we can easily find that $x+y=a^2-b^2=pqrs$ so \begin{align*} x&=\frac{-pq+rs+pqrs}{2}\\ y&=\frac{pq-rs+pqrs}{2} \end{align*}Then using $(2x)(2y)=(2a)^2$ we can get $(pqrs)^2=(pr+qs)^2+(pq-rs)^2=(p^2+s^2)(q^2+r^2)$. We have either $p^2s^2\le p^2+s^2$ or $q^2r^2\le q^2+r^2$, which implies $\min{p,q,r,s}=1$. WLOG, assume $p=1$ then \[s^2q^2r^2=(1+s^2)(q^2+r^2)\le 2s^2(q^2+r^2)\]so $q^2r^2\le 2(q^2+r^2)$ and it is clear at this point that $s=1$ and $q=r=2$ which gives us the values $x=y=2$ and $z=4,t=0$ which is absurd.

The answer is no. Let $a=\tfrac 12 (x+y)=\tfrac 12 (z+t)=\tfrac 12(xy-zt)$; let $x = a - \alpha$, $y = a+\alpha$, $z = a-\beta$, $w = a+ \beta$. Without loss of generality, $\alpha,\beta>0$. Suppose to the contrary that $xy$ and $zt$ are perfect squares; write $xy = \gamma^2$, $zt = \delta^2$. Then \begin{align}\tag{1} xy - zt = (a^2 - \alpha^2)- (a^2-\beta^2) = \beta^2 - \alpha^2 = 2a. \end{align}Additionally, we have \begin{align}\tag{2} a^2 = (a^2 - \alpha^2) + \alpha^2 = \gamma^2 + \alpha^2=\delta^2+\gamma^2 \end{align}Lemma 1. $a\in \mathbb{Z}$. Proof. If not, then $a = a_0 + \frac{1}{2}$ for $a_0\in\mathbb{Z}$; then since $x=a-\alpha\in\mathbb{Z}$, we have $\alpha = d_1 + \frac{1}{2}$ for $d_1\in \mathbb{Z}$, similarly $\beta = d_2+\frac{1}{2}$ for $d_2\in \mathbb{Z}$. Then note that \begin{align} \alpha^2 = (d_1+\tfrac 12)^2 = \underbrace{d_1(d_1+1)}_{\text{even}}+\tfrac{1}{4} \end{align}and the corresponding equation for $\beta^2$, from which it is evident that $2a_0+1 = 2a = \beta^2 - \alpha^2$ is even, a contradiction. $\blacksquare$ Since $2a\in \{0,2\}\pmod 4$ and $\beta^2 - \alpha^2 \in \{-1,0,1\}\pmod 4$ it follows that $a = 2a_1$, for $a_1\in \mathbb{Z}$. Then since $\gamma^2 + \alpha^2\equiv 4a_1^2 \equiv 0\pmod 4$, we get $\gamma = 2\gamma_{1}$ and $\alpha= 2\alpha_{1}$ and similarly $\delta = 2\delta_1$, $\beta = 2\beta_1$ for $\alpha_1,\ldots, \delta_1\in \mathbb{Z}$. Thus (1) and (2) become \begin{align}\tag{$1'$} \beta_{1}^2 - \alpha_{1}^2 &= \frac{a_1}{2} \\ \tag{$2'$} \gamma_{1}^2 + \alpha_1^2 &=\delta_1^2 + \beta_1^2= a_1^2. \end{align}Lemma 2. For positive integers $p_{1},p_2, q_1,q_2, r,k$, if we have \begin{align}\tag{$\dag$} p_2^2 -p_1^2 = \frac{r}{2^k} \\ \tag{$\ddag$} p_i^2 + q_i^2 = r^2, \end{align}then $2\mid p_i,q_i,r$. Proof. Clearly $2^k\mid r$, so $2\mid r$. Then $p_{i}^2 + q_i^2 \equiv r^2 \equiv 0\pmod 4$, so we must have $2\mid p_i,q_i$. $\blacksquare$ Then, writing $\widetilde{p}_i = p_i/2$, etc. and substituting into ($\dag$) and ($\ddag$) we get \begin{align}\tag{3} \widetilde{p}_2^2 -\widetilde{p}_1^2 &= \frac{\widetilde{r}}{2^{k+1}} \\ \tag{4} \widetilde{p}_i^2 + \widetilde{q}_i^2 &= \widetilde{r}^2, \end{align}So we can apply the lemma again to get $2\mid \widetilde{p}_i, \widetilde{q}_i,\widetilde{r}$. Continuing this process yields $2^n\mid p_i,q_i,r$ for any $n>0$, which is obviously impossible. But since $\alpha_1,\beta_1,\gamma_1,\delta_1, a_1$ satisfy the conditions of the lemma, we have a contradiction, so we are done.

Did not enjoy. Answer: No. Suppose yes. Let $p^2=xy$ and $zt=q^2$. Relabel so that $z$ is now $y$. Then our condition becomes \[p^2-q^2=\frac{p^2+x^2}{x}=\frac{q^2+y^2}{y}\]Turning these into quadratic equations we get $x^2-x(p^2-q^2)+p^2=0$ $y^2 - y(p^2-q^2) + q^2 = 0$ For these to have integer roots both $(p^2-q^2)^2-4p^2$ and $(p^2-q^2)^2 - 4q^2$ need to be perfect squares. Now, note that if $2 < p \neq q + 1$ \begin{align*} \left(\frac{p^2-q^2}{2} - 2\right)^2 &\leq (p^2-q^2-2p)^2 =(p^2-q^2)^2-4p(p^2-q^2)+4p^2 \leq \\ &\leq (p^2-q^2)^2-4p(4p-4)+4p^2 = (p^2-q^2)^2 - 12p^2 + 16p < (p^2-q^2)^2 -4p^2 \end{align*}Then, letting $x^2=(p^2-q^2)^2-4p^2$, we notice that \[(x+4)^2-x^2 > \left(\frac{p^2-q^2}{2}+2\right)^2 - \left(\frac{p^2-q^2}{2}-2\right)^2 = 4(p^2-q^2) \implies (x+4)^2 > (p^2-q^2)-4q^2\]Hence, $(x+2)^2=(p^2-q^2)^2-4q^2$, by parity. This implies $(x+2)^2-x^2=2(2x+2)=4(p^2-q^2)$ and hence $x=p^2-q^2-1$. This obviously cannot be true since it would imply that $x^2$ is not congruent to $(p^2-q^2)^2-4p^2$ modulo $2$. Now, suppose that $p=q+1$. Then, we would have $(p^2-q^2)^2-4p^2 < 0$ which implies that $x$ does not exist, a contradiction. $\blacksquare$

Goofy non-NT solution

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