My (one line ) solution:

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TST? oh it's also for iran (95 or 96) see here : http://www.mathlinks.ro/Forum/viewtopic.php?p=204325&highlight=

Another solution: if $3x^2+x=4y^2+y$, then $(3x+3y+1)(x-y)=y^2$. Now if $p$ is a prime and $p \mid x-y$, we have $p \mid y^2 \Rightarrow p \mid y \Rightarrow p \mid x-y+y=x$. Now suppose $p \mid x-y, p \mid 3x+3y+1$, then as $p \mid x, p \mid y$ we have $p \mid 3x+3y-1-3x-3y=1$, contradiction, so $(3x+3y+1, x-y)=1$. Thus the result.

Also from the same contest: Find all numbers x such that both x and x inverse ( x in reverse order) are powers of 2. (same level of difficulty)

ali wrote: Also from the same contest Which one? Not the french TST... Please start a new thread whith this new problem. Pierre.

Igor wrote: My (one line ) solution:

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Nice solution. Actually we can prove that $x-y = (\gcd (x,y))^2 $

the relation can be developped and so on: (x-y)(1+3(x+y))=y² note a=(x-y) and b=(x+y) then 4a(1+3b)=(b-a)² the equation X²-14aX+(a²-4a)=0 has b as a solution so the small discriminant is a perfect square this discriminant =48a²+4a=4a(12a+1) so 4a|a perfect square and gcd((12a+1),4a)=1 so 4a is perfect square

We can rewrite $ 3x^2 + x = 4y^2 + y$ as: $ (8x + 1)^2 = (4x)^2 + (8y + 1)^2$ but that is no more than a pythagorean triple. Exist $ u,v$ positive integers whit some conditions such that: $ u^2 + v^2 = 8x + 1$ $ 2uv = 4x$ $ u^2 - v^2 = 8y + 1$ because $ 4x$ is even, while $ 8y + 1$ is odd. From the first and the third ecuation: $ 2v^2 = 8(x - y)$, then ($ 2 \mid v$ and) $ (\frac {v}{2})^2 = x - y$