Can an infinite set of natural numbers be found, such that for all triplets $(a,b,c)$ of it we have $abc + 1 $ perfect square? (20 points )
Problem
Source: Iranian 3rd round Number Theory exam P5
Tags: ratio, algebra, polynomial, number theory proposed, number theory
individ
22.09.2014 08:49
I think the problem must be corrected. Because when this statement is trivial. Because it is equivalent to: $abc+1=n^2$ $cab=(n-1)(n+1)$ For a given number can always be decomposed into 3 multiplier. It is necessary that one number was specified and you need to find the rest. And let set will be: $c$
individ
22.09.2014 09:33
At first when I started to solve the equation of thought that you can specify only one factor. $c$ Were you can ask any ratio $a$ And the solution of the equation: $cab+1=n^2$ $b=k(ack+2)$ $n=cak+1$ $k$ - any integer.
math-helli
24.09.2014 12:38
This beautiful problem is designed by Mohsen Jamaali our great teacher in number theory
v_Enhance
27.09.2014 19:51
individ wrote: For a given number can always be decomposed into 3 multiplier. This doesn't make sense. The problem is asking whether $abc+1$ is a square for any $a$, $b$, $c$ of some infinite set $S$, not whether every perfect square can be written as $abc+1$. Presumably the problem also requires $a$, $b$, $c$ to be distinct, because otherwise Mihăilescu's theorem kills it. Would love to see a solution to this, the problem looks quite nice indeed.
individ
27.09.2014 20:59
v_Enhance wrote: individ wrote: For a given number can always be decomposed into 3 multiplier. This doesn't make sense. The problem is asking whether $abc+1$ is a square for any $a$, $b$, $c$ of some infinite set $S$, not whether every perfect square can be written as $abc+1$. Presumably the problem also requires $a$, $b$, $c$ to be distinct, because otherwise Mihăilescu's theorem kills it. Would love to see a solution to this, the problem looks quite nice indeed. I didn't understand what you want? The formula you wrote, what else do you need? If you want numbers were excellent, you can choose different. What number do You have problems?
ZetaX
27.09.2014 23:55
Your "solution" is gibberish. You probably missunderstood the problem completely. Read it again. Anyway, the problem should have added that $a \neq b \neq c \neq a$ as otherwise it is quite easy.
individ
28.09.2014 07:53
What else can be complicated? If: $n=17$ $cab=(17-1)(17+1)=16*18=4*8*9$ These numbers different? Or another: $cak=n-1=16=2*8*1$ $b=16+2=18$ What? This method always get different numbers. What else is needed?
mavropnevma
28.09.2014 13:07
ZetaX wrote: Your "solution" is gibberish. You probably missunderstood the problem completely. Read it again. Anyway, the problem should have added that $a \neq b \neq c \neq a$ as otherwise it is quite easy. A way to avoid that is to use triplets $\{a,b,c\}$ rather than $(a,b,c)$, as the use of the set notation implicitly implies $a,b,c$ to be pairwise distinct
ZetaX
28.09.2014 21:53
mavropnevma: I am not so sure that this implicitely implies them to be distinct. It is probably a convention, but I am pretty sure there are people using it differently. For example such a thing as "if $p$ is not an element of $\{2,q\}$", where $p,q$ are primes. individ: an infinite set $S$ of numbers such that for all distinct $a,b,c \in S$ we get $abc+1$ to be a square. You just gave an infinite amount of sets with three elements.
dibyo_99
07.10.2014 07:02
Suppose that an infinite set $S$ with the desired property does exist. Fix any $3$ elements $a, b, c$ in the set. Then $abx+1$, $bcx+1$, $cax+1$ are all perfect squares $\forall x \in S - { a,b,c } $. Then, the equation $y^2 = (abx+1)(bcx+1)(cax+1)$ has infinitely many solutions over integers, contradicting Siegel's Theorem on Elliptic Curves.
Suppose that an infinite set $S$ with the desired property does exist. Fix any $4$ pairwise distinct elements $a< b < c < d$ in the set. Then, $abx+1, bcx+1, cdx+1, dax+1$ are all perfect squares $\forall x \in S - { a,b,c,d }$. Then, the polynomial $P(x) = (abx+1)(bcx+1)(cdx+1)(dax+1)$ takes over perfect square values $\forall x \in S - { a,b,c,d } $. Check out olorin's post here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=148696 Since $P$ has even degree $=4$ and its leading coefficient $= a^2b^2c^2d^2$ is a perfect square, it must be the square of some integer polynomial $Q(x)$. Since $P$ is biquadratic, so $Q$ must be a quadratic. Now, we have rather few choices for $Q$. It must be the product of some $2$ factors of $P$, and, the product of the other $2$ must also be equal to $Q$.
If $abx+1$, is paired with $bcx+1$, we have $(abx+1)(bcx+1) = (cdx+1)(dax+1)$. Comparing coefficients of the $x^2$-term, we have $ab^2c = cd^2a \implies b^2 = d^2$, contradicting $d > b$.
If $abx+1$ is paired with $dax+1$, we have $(abx+1)(dax+1) = (bcx+1)(cdx+1)$. Comparing coefficients, $a^2bd = bc^2d \implies a^2 = c^2$, contradicting $c > a$.
If $abx+1$ is paired with $cdx+1$, we have $(abx+1)(cdx+1) = (bcx+1)(dax+1) \implies ab+cd = bc+da$. Factoring, we get $(b-d)(a-c) = 0$. Therefore, $a=c$ or $b=d$, contradicting the fact that they are pairwise distinct.
I'd be stuck with the overkill, if it weren't for you, v_Enhance
v_Enhance
09.10.2014 01:45
Well, math154 told me about that solution, so that's probably who you want to thank Apparently this problem was on the last Black MOP test of MOP 2013, which implies that it's not new.