a) This statement is false. For example, $(a,b,c) = (1,3,8)$ yields $(ab+1,bc+1,ca+1) = (2^2,5^2,3^2)$.

Here is the official wording Quote: a) Arătaţi că, oricare ar fi a, b, c, \in N , numerele ab+1,bc+1 si ca+1 nu pot fi simultan pătrate perfecte pare. b) Arătaţi că există o infinitate de numere naturale distincte două câte două a, b, c şi d, astfel încât numerele ab+1,bc+1, cd+1 şi da+1 să fie simultan pătrate perfecte

For part $b$ the quadruple $\{1,n^2-1,n^2+2n,n^2+4n+3\}$ yields $\{n^2,(n^2+n,-1)^2,(n^2+3n+1)^2,(n+2)^2\}$.

https://artofproblemsolving.com/community/c3046h1184227_the_system_is_an_arithmetic_progression https://artofproblemsolving.com/community/c3046h1183749_system_with_permutations https://artofproblemsolving.com/community/c3046h1057324_the_system_is_almost_linear_diophantine_equations

In a) asks:"... can these numbers be simultaneously even perfect squares". We have that $ab, bc, ac=-1(mod4)$ and thus $(abc) ^2=-1(mod4) $, impossible.

the original incorrect wording was this parmenides51 wrote: a) Show that, for any $a, b, c, \in N$, the numbers $ab+1,bc+1$ and $ca+1$ can not be simultaneously perfect squares. b) Show that there is an infinity of natural numbers (distinct two by two) $a, b, c$ and $d$, so that the numbers $ab+1,bc+1, cd+1$ and $da+1$ are simultaneously perfect squares. therefore that reply was also correct Solar Plexsus wrote: a) This statement is false. For example, $(a,b,c) = (1,3,8)$ yields $(ab+1,bc+1,ca+1) = (2^2,5^2,3^2)$. I am correcting the wording in part (a), according to @above's reply to parmenides51 wrote: a) Exist $a, b, c, \in N$, such that the numbers $ab+1,bc+1$ and $ca+1$ are simultaneously even perfect squares ?

a) The perfect squares seem to be multiples of 4. In this case we would have $ab, bc ,ac \equiv -1 (mod 4)$, so $(abc)^2=(ab)(bc)(ca) \equiv -1 (mod 4)$, which is absurd. b)Taking $d = 0$, we have $cd + 1 = da + 1 = 1^2$ for any $a, b, c$, so it is sufficient to find distinct non-zero natural numbers $a, b, c$ such that $ab + 1$ and $bc + 1$ are simultaneously perfect squares. Let us then take $a = n - 2, b = n$ and $c = n + 2$, for natural $n > 2$. We will have $ab + 1 = (n - 2) n + 1 = (n - 1)^2$ and $bc + 1 = n(n + 2) + 1 = (n + 1)^2$. .