Can someone help me to prove that is not divisible by 121 for any ?
Suppose that for some that is divisible byOriginally Posted by OReilly
then the exists a such that:
So from the quadratic formula we have:
Which implies that is an odd integer, say. So:
but the LHS is divisible by , and as is a square it is also divisible
by (which is ), but that would imply that is divisible by ; a
contradiction, so no such exits.
RonL
If is divisible by 121 then so is . If is divisible by 121 then it's divisible by 11 and hence 11 divides . If 11 divides a perfect square then it divides : and that implies that 121 divides . So we have 121 dividing and , hence the difference, which is 11. But that's a contradiction.Can someone help me to prove that is not divisible by 121 for any ?
If a square is divisible by a prime, it is also divisible by the square of theCan you just explain me this part, I didn't quite understand it.
Originally Posted by CaptainBlack
prime. It's a consequence of the fundamental theorem of arithmetic, that
the decomposition of a number into a product of primes is essential unique.
So the left hand side of the equation being divisible by 11 implies that the
right hand side is also divisible by 11, but the right hand side is a square.
So the right hand side is divisible by 121 (the square of 11). This implies
that the left hand side is divisible by 121, and so that 11 is divisible by
121 which is a contradiction.
RonL
When I was trying to solve the problem I have done this:
But I have stoped there.
Let me see if I have understood good.
is obviosly divisible by 11, so is also divisible by 11. But is also divisible by which now implies that is divisible by 121. So we get that which shows that 11 is not divisible by 121.
Is that correct?
I didn't know that if a square is divisible by a prime, it is also divisible by the square of the prime. Boy, I feel stupid and embarrassed now!
Divisibility of numbers is simply killing me, I just can't solve any harder problem. I am not expert, but I think that in order to solve those problems you need to know Number theory. Since I am on level of Algebra 1 and 2 for high school (problem that I have posted is from Algebra 1) I find that many harder problems constructed for learning Algebra 1 and 2 that concerns divisibility are based on writers thorought knowledge of Number theory. I don't know if I am wrong, but that's my impression. Either that, or I must find excuse!
Proof,Originally Posted by OReilly
Let then, or . If then, then, by Euclid's Lemma. Thus, .
Thus, given you have, thus, or thus, definetly. Now, if then, but since then if you want to be formal about it
That is right keep on feeling stupid.Originally Posted by OReilly
I don't see how this last equation follows from what you say it does.When I was trying to solve the problem I have done this:
But I have stoped there.
Let me see if I have understood good.
is obviosly divisible by 11, so is also divisible by 11. But is also divisible by which now implies that is divisible by 121. So we get that
RonLwhich shows that 11 is not divisible by 121.
Is that correct?
I didn't know that if a square is divisible by a prime, it is also divisible by the square of the prime. Boy, I feel stupid and embarrassed now!
My lack of understanding again!Originally Posted by CaptainBlack
Let me try again.
From my example we have:
That implies that is divisible by 11. But is also divisible by 121 but we would have that is also divisble by 121, which is different.
Is that ok?