Question 34722This question is from textbook
: Prove with a simple equation that the product of four consecutive integers can never be a perfrct square.
This question is from textbook
Found 2 solutions by longjonsilver, venugopalramana:
Answer by longjonsilver(2297)   (Show Source):
You can put this solution on YOUR website! not sure if this is a rigourous proof, but it should at least show you some things.
Take an example: 2*3*4*5. To be a perfect square, we need something like 3*3 or 7*7 or 15*15 etc.
So with 2*3*4*5, we need two of the integers to multiply to equal the other two integers. The only remote possibility of this is if 2*5=3*4, which it doesn't in this case, since 10 is not 12.
So, algebraically, x(x+1)(x+2)(x+3) are our four integers, starting at number "x".
If we then say, OK, for which value of x does x(x+3) = (x+1)(x+2) hold? then we can find the answer(s). From the wording of the question, we are looking for no solution.
which is patently not true, so there is no value of x such that is true.
jon.
Answer by venugopalramana(3286)  (Show Source):
You can put this solution on YOUR website! LET THE 4 CONSECUTIVE INTEGERS BE N-1,N,N+1,N+2
THEIR PRODUCT = P =(N-1)N(N+1)(N+2)=(N^2-1)(N^2+2N)=N^4+2N^3-N^2-2N....IF THIS IS TO BE A PERFECT SQUARE THEN IT SHOULD BE IN THE FORM OF
(N^2+AN)^2...SINCE THERE IS NO CONSTANT TERM IN P...EXPANDING
(N^2+AN)^2=N^4+2AN^3+A^2N^2..EQUATING WITH P WE SHOULD HAVE
2A=1...OR...A=1/2
A^2=-1..AND -2N=0.....WHICH IS NOT POSSIBLE .....HENCE P CANNOT BE WRITTEN AS A PERFECT SQUARE.
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