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Are there any solutions to: x!y!=z! for any x,y and z.
The number [100a+10b+c] is a prime. Prove that b^2-4ac cannot be a perfect square. Solomon7968 (talk) 11:06, 21 March 2013 (UTC)[reply]
1) Allowing some of x, y, z to be 0 or 1 gives lots of solutions. Otherwise I don't think it's possible. Staecker (talk) 11:36, 21 March 2013 (UTC)[reply]
1) There's also a family of solutions (n!-1)!n! = (n!)! e.g. 5!3! = 6!. Gandalf61 (talk) 12:40, 21 March 2013 (UTC)[reply]
Welcome to Wikipedia. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our aim here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. See further our articles on factorial and quadratic formula. Pokajanje|Talk 15:51, 21 March 2013 (UTC)[reply]
Are x, y, and z supposed to be natural ? Either way, you might also want to check this out. Also, are a, b, and c supposed to be digits from 0 to 9 ? Frankly, I see no reason why b2 - 4ac can not be a perfect square... all that is certain is that b2 - 4ac + 4ap is supposed to be one, and the two conditions do not seem to be mutually exclusive... — 79.113.245.81 (talk) 21:52, 21 March 2013 (UTC)[reply]
There is a relatively easy non-trivial set of solutions to the factorials problem. To find it, you must remember that n! = n⋅(n-1)!.
There are, however, solutions outside this particular set (but which follow a similar logic). It burned my brain to find one, but I did: 6!7!=10! because 6!=720=8×9×10. Thank you and good night.
Oh, and if you use advice from the reference desk for HW at the university level, you must cite it on your paper (as with any person irl from whom you received help). Not doing so may be considered bad. 14:52, 22 March 2013 (UTC) (By User:SamuelRiv)
Another solution set can be found by setting y=z-2 and solving the quadratic eqn for z in terms of x (this method shown to me by a colleague). Then you get a condition for x alone (which may be related to Brocard's problem, suggesting finitely many solutions). The smallest solution I found via plug-and-chug was x=19, giving z=348776557. SamuelRiv (talk) 17:41, 23 March 2013 (UTC)[reply]
Re second question: a=0, b=2, c=3. [100a+10b+c]=23, a prime. b^2-4ac = 2^2-4*0*c = 4, a square number. Did I misunderstand the question? --NorwegianBluetalk 21:13, 22 March 2013 (UTC)[reply]
Given the way that the problem is written, I presumed that 'a' had to be greater than zero. Just because we've probably gone passed when the HW was due, I think the major points in the proof are something like
if b^2+4ac is a perfect square then (-b+/-sqrt(b^2-4ac))/2a has two integer values, m & n.
In that case the factorization of ax^2+bx+c is of the form x-m and x-n, which means that a*10^2+b*10+c can be factored into 10-m and 10-n.
The piece that needs to be handled is making sure that for these values that m(or n) can't be 9 which would allow for a factorization, but would still allow b^2+4ac to be a square. I'm not quite sure how to do that.Naraht (talk) 21:41, 24 March 2013 (UTC)[reply]
