\hspace{-15}$\textbf{Find Integer solution for}\\\\ $\mathbf{x(x+1)(x+2)(x+3)(x+4)(x+5)=y^2-1}$
21This is going to be easy..
hint: the gap between two perfect squares...
11Suppose y^2\ne1.
Then take x^2+5x+2=k
So now the eqn becomes
(k-2)(k+2)(k+4)=y^2-1
Recall that the product of r consecutive positive integers is divisible by r!
So 720|(y^2-1)\Rightarrow (y^2-1)\equiv 0(mod4)
That also gives 4|k\Rightarrow 4|(x^2+5x+2)
From which we arrive at (x^2+5x+2)=\pm 8
The solutions of which turn out to be (-6,1)
None of which satisfy the eqn. So y^2=1
1708Thanks shumik and shubhopip.
can anyone give me a link of modulo theory for polynomial.
thanks.
11Srry - neglect the above reply, it's flawed. Looks like I was completely out of my mind while posting it.
3if we put x = 0 then RHS = 0 and to make LHS 0 we can put y as +1 and -1.
also LHS is 0 at x = -5, -4 ,-3, -2 and at -1.
at all these values RHS = 0 at y +1 and -1 .
11Why do you guys go on blindly guessing solutions?
A nice bound is attainable for x, because after some k, the prod. on the lhs is divisible by 100.
Now the important step is to notice the last non-zero digits.
The eqn infact can be re-stated as k!+(k-6)!=(k-6)!y^2
Now observe the last non-zero digit of (k-6)! becomes periodic, of the order (8,8,6,8,2), after which this sequence again reccurs.
Now comparing these, we can easily get a contradiction to say the least that the given prod. cannot be divisible by 100. (Check, and prove this, the hint is to check the last non-zero digit on the two sides, given y2≡01(mod100))
So the bound for k (and in turn x) is easily attainable.
And for those trying to make both the sides of the given eqn 0, my sincere apologies, y=±71 is also a soln.
21Ignore my 1st post.. i saw something else...:P
man111 wrote ''Thanks shumik and shubhopip''
ROFL!!!
