(72, 4, x²-4, 1)

Such a trivial example. Just take (a; a+2; a+10; a+6)

$(9,32,4,k^2-1)$

The most trivial out of all $(0,0,0,k)$ where $k$ is an ineteger.

ohhh wrote: (72, 4, x²-4, 1) Enigma714 wrote: $(9,32,4,k^2-1)$ DensSv wrote: The most trivial out of all $(0,0,0,k)$ where $k$ is an ineteger. Are these solutions allowed? Can we fix a,b,c or d?

DensSv wrote: The most trivial out of all $(0,0,0,k)$ where $k$ is an ineteger. Actually I should’ve written positive integers. Mistake already corrected

Take $(a,b,c,d)=(k-6,k-4,k+4,k)$. We can check that these work. $\square$

alexgsi wrote: ohhh wrote: (72, 4, x²-4, 1) ... Are these solutions allowed? Can we fix a,b,c or d? Don't think so. The problem is not requesting infinitely many tetrads $(a, b, c, d)\in \mathbb{Z}^{+}$ but "infinitely many positive integers $a, b, c$ and $d$" (which is a short way of saying "infinitely many $a \in \mathbb{Z}^{+}$, infinitely many $b \in \mathbb{Z}^{+}$, infinitely many $c \in \mathbb{Z}^{+}$ and infinitely many $d \in \mathbb{Z}^{+}$).

coquitao wrote: alexgsi wrote: ohhh wrote: (72, 4, x²-4, 1) ... Are these solutions allowed? Can we fix a,b,c or d? Don't think so. The problem is not requesting infinitely many tetrads $(a, b, c, d)\in \mathbb{Z}^{+}$ but "infinitely many positive integers $a, b, c$ and $d$" (which is a short way of saying "infinitely many $a \in \mathbb{Z}^{+}$, infinitely many $b \in \mathbb{Z}^{+}$, infinitely many $c \in \mathbb{Z}^{+}$ and infinitely many $d \in \mathbb{Z}^{+}$). The problem can be validly interpreted as infinitely many ordered quadruplets $(a,b,c,d)$ satisfying the thing If it's asking for infinitely many $a$, infinitely many $b$, ... it should say that

coquitao wrote: alexgsi wrote: ohhh wrote: (72, 4, x²-4, 1) ... Are these solutions allowed? Can we fix a,b,c or d? Don't think so. The problem is not requesting infinitely many tetrads $(a, b, c, d)\in \mathbb{Z}^{+}$ but "infinitely many positive integers $a, b, c$ and $d$" (which is a short way of saying "infinitely many $a \in \mathbb{Z}^{+}$, infinitely many $b \in \mathbb{Z}^{+}$, infinitely many $c \in \mathbb{Z}^{+}$ and infinitely many $d \in \mathbb{Z}^{+}$). The problem creators affirmed that they meant quadruplets.

$(a, b, c, d) = (k+10, k+8, k, k+4)$ works for all $k$ positive integer: $ab+1 = (k+10)(k+8) + 1 = k^2 + 18k + 81= (k+9)^2$ $bc+16 = (k+8)(k) + 16 = k^2 + 8k + 16= (k+4)^2$ $cd+4 = (k)(k+4) + 4 = k^2 + 4k + 4= (k+2)^2$ $da+9 = (k+4)(k+10) + 9 = k^2 + 14k + 49= (k+7)^2$ Motivation: It's natural to think how could I find a perfect square $ab+1$ , as a first step. Then, we notice that $(w-1)(w+1) + 1 = w^2 -1 +1 = w^2$ Similarly: $(x-4)(x+4) + 16 = x^2$ $(y-2)(y+2) + 4 = y^2$ $(z-3)(z+3) + 9 = z^2$ Now, we need the equalities and, considering $3 - 1 - 4 + 2 = 1 + 4 - 2 - 3$, we are going to search $w, x, y, z$ positive integers such that: $w-1=x+4$ $x-4=y-2$ $y+2=z-3$ $z+3=w+1$ So $(w, x, y, z)=(y+7, y+2, y, y+5)$ is a solution. (Use the distance between the numbers to find the solutions). Then: $(y+6)(y+8) + 1 = w^2$ $(y-2)(y+4) + 16 = x^2$ $(y-2)(y+2) + 4 = y^2$ $(y+2)(y+8) + 9 = z^2$ Using $y-2=k$: $(k+8)(k+10) + 1 = w^2$ $(k)(k+8) + 16 = x^2$ $(k)(k+4) + 4 = y^2$ $(k+4)(k+10) + 9 = z^2$

Let: $ab+1=r^2$ $bc+16=q^2$ $cd+4=s^2$ $ad+9=p^2$ for positive integers $r,s,p,q$ $\implies r-1=a, r+1=b \implies b=a+2$ $q-4=b, q+4=c \implies q-4=a+2 \implies c=a+10$ $p-3=a, p+3=d \implies d=a+6$ So we have $(a,a+2,a+10,a+6)$ has a solution and testing it: $a(a+2)+1=a^2+2a+1=(a+1)^2$ $(a+2)(a+10)+16=a^2+12a+36=(a+6)^2$ $(a+10)(a+6)+4=a^2+16a+64=(a+8)^2$ $a(a+6)+9=a^2+6a+9=(a+3)^2$ so it works. Just let $a=n$ and we'll have the infinite set of solutions $(a,b,c,d)=(n,n+2,n+10,n+6)$ where $n$ is a positive integer and we're done.

Yet another systematic way: we search $a,b,c,d$ such that $ab+1=x^2$, $bc+16=y^2$, $cd+4=z^2$ and $da+9=t^2$. Notice if $ab=x^2-1$ then the simple choice $(a,b)=(x-1,x+1)$ gives $a-b\in\{\pm 2\}$. Call $\Delta(a,b):=a-b$. We have $\Delta(a,b)=\pm 2$, $\Delta(b,c)=\pm 8$, $\Delta(c,d)=\pm 4$ and $\Delta(d,a) = \pm 6$. Furthermore, $\Delta(a,b)+\Delta(b,c)+\Delta(c,d)+\Delta(d,a)=0$ must hold. It is very easy to see that the choice $\Delta(a,b)=2$, $\Delta(b,c)=8$, $\Delta(c,d)=-4$ and $\Delta(d,a)=-6$ satisfies the conditions. Equipped with this, we choose $a=x+1$ and $b=x-1$. Next, we choose $b=y+4$ and $c=y-4$, that is, $y=x-5$. Likewise, $c=z-2$ and $d=z+2$, which is satisfied for $z=x-7$. Finally, $d=x-5$ and $da+9 = (x-5)(x+1)+9 = (x-2)^2$. So, $(a,b,c,d)=(x+1,x-1,x-9,x-5)$ works for all $x$.

$(a, a+2, a+10, a+4) $ works for all $a \in \mathbb{N}$.

Any quadruplet $(a,b,c,d)=(p+1,p-1,p-9,p-5)$ where $p \in \mathbb{N}$ works.

Let $(a,b,c,d)=(a, a-2, a-10, a-6)$, then we have: $$ab+1=a(a-2)+1=a^2-2a+1=(a-1)^2$$$$bc+16=(a-2)(a-10)+16=a^2-12a+36=(a-6)^2$$$$cd+4=(a-10)(a-6)+4=a^2-16a+64=(a-8)^2$$$$da+9=a(a-6)+9=a^2-6a+9=(a-3)^2$$Therefore, for every $a\geq11$, there exists a solution.

(k, k+2, k+10, k+6)

thdwlgh1229 wrote: $(k, k+2, k+10, k+6)$