The excitement you embody when you say 'Do you think this is a perfect square' is exactly the excitement that anybody should feel when they get to that point. You humanize all that is good with math!
At the point at which you’re looking at x^4 + 6x^3 + 11x^2 + 6x + 1, I would be tempted to “try” (x^2 + cx + 1)^2 and solve for c if it looks promising. We expect a square, by the way the question is posed. Also, all coefficients are positive, so a little thought tells us we can’t have any negative coefficients in the factor, so we don’t have to try -1 (or -x^2). This gives us x^4 + 2cx^3 + (c^2 + 2)x^2 + 2cx + 1, immediately giving 2c=6, c=3, which thankfully is compatible with c^2+2=11. So the guess is right, and we just plug 96 into x^2+3x+1
I did it an unconventional way (and I don't know if it's "legal" in maths) : I've tried with x=1, and the result was 25, 5², or (1*4 +1)² Then I tried with x=2, and the result was 121, 11² or (2*5 +1)² So I theorized the result could be (x*(x+3) + 1)² and I wrote the calcul : x(x+1)(x+2)(x+3) + 1 = (x(x+3) + 1)^2 I developped both sides and got x^4 + 6x^3 + 11x^2 + 6x + 1 = x^4 + 6x^3 + 11x^2 + 6x + 1 The same calcul on both sides, so my assumption seems correct. Put under the square root, it gives x(x+3) + 1. Replace x with 96, I have 96*99 + 1, or 9504 + 1 or 9505. Edit : minor corrections
I gave this one a try and set x to be the second number of the four, because (x-1)(x+1) was a nice, easy product to deal with. The factoring was a bit guess-and-check, but since it started with x^4 and ended with 1, the square root was obviously x^2+ax(+/-)1, and the coefficients I had were all ones or twos, so I only had four meaningful options for a. The end result was x^2+x-1, which looks like a nice, neat answer.
My way for this state is: Suppose those numbers ( 99,96,97,98 )that are 100 So sq(100^4) is 100^2=10000 But we know those numbers are not really 100 ,they less &soon Then the answer is less&soon 10,000
I liked your approach but I was tempted by how close the numbers were to 100 this is my attack on this problem sqrt(96*97*98*99 + 1) this is actually a bit simpler if you take h = 100 and substitute: sqrt((h-4)(h-3)(h-2)(h-1) + 1) then switch the pairs to make pairs with numbers that sum up to 5 sqrt( (h-4)(h-1) * (h-3)(h-2) + 1 ) then multiply the pairs out sqrt( (h^2 - 5h + 4)(h^2 - 5h + 6) + 1 ) then we notice the first two terms of each polynomial in the product are the same (h^2 - 5h) so substitute n = h^2 - 5h which substituting back 100 for h works out to n = h^2 - 5h = (100^2) - (5*100) = 10000 - 500 = 9500 so actually n = 9500 putting n in this expression works out to sqrt((n+4)(n+6) + 1) multiply the binomials out sqrt(n^2 + 10n + 24 + 1) adding the two constants sqrt(n^2 + 10n + 25) now we see that this is equivalent to sqrt(n^2 + 2 * 5 * n + 5^2) so we actually have a square of a binomial sqrt( (n + 5) ^ 2 ) the square and the square root cancel out n + 5 substitute back n = 9500 which we figured out when getting n from h^2 - 5h 9500 + 5 which makes the answer 9505
I was thinking of the product of sums for weighted choice of x at some middle coefficient to let x = 97 or x = 98 but that fails for even multiplied x sums. So this method turns out better. Plus when I divided out the x^2 to get an answer I forgot to remuktiply it back in. I was doing the solution in R^2 and then the multiplied 11 coefficient dividing out by x^2 I just ignored that in finding the final results I saw you get with R^2 = x^2 multiplied back in. I need this brainwork practicing because I do mathematics statements errors like that too often! 😬
If you take the absolute value, you get a valid expression of R(x)= | x²+3x+1 | = √(x.(x+1).(x+2).(x+3)+1) for any positive, null or negative real x. The square root is always defined since the palindromic polynômial x⁴+6x³+11x²+6x+1 is never negative whatever the value of x. But the absolute value is mandatory in R(x) expression since polynômial x²+3x+1 is négative when x is in ] (-3-√5)/2 ; (-3+√5)/2 [. Note that R(x)=1 for x in { -3 -2 -1 0 } only. Furthermore R(x)=0 at x=(-3±√5)/2 respectively x ≈ -2.6180 and x ≈ -0.38197 next interesting problem is to found how positive and negative integers p>0 and n
The formula works for all strictly positive integer. Does it works for 0 ? well, strangely enough yes, although, the proof doesn't work if x = 0 What about negative numbers ? for -3, -2, -1 and 0, R is 1, no need for a formula for any number below -3, we are multiplying 4 negative numbers, which is the same as multiplying 4 positive numbers, but |x| is now the biggest of the 4 so with y = -(x+3) R = y^2 + 3y +1 or R = (x+3)^2 - 3x -8
Palindromic Property is a little complicated to me , so i try this one , lucky , it works. (x-d)xy(y+d)+d^4 where x+d=y =(x^2-xd)(y^2+yd)+d^4 =(xy)^2+x^2yd-xy^2d-xyd^2+d^4 =(xy)^2-xyd(y-x)-xyd^2+d^4 =(xy)^2-2xyd^2+d^4 =(xy-d^2)^2
If you multiply x with x+3 and x+1 with x+2 then you can say t= x²+3x and it will be even more easier to calculate
shouldn't you add the 1 to get t = x²+3x + 1 ?
yes yes yes very goooood. I agree with this opinion
It will be even more easy if we write it sqrt(100-4)(100-3)(100-2)(100-1)
@@marineintelligence you can do that
Yes, it should be done this way so that it showed that the product is 1 less than a perfect square
The excitement you embody when you say 'Do you think this is a perfect square' is exactly the excitement that anybody should feel when they get to that point. You humanize all that is good with math!
Really beautiful how numbers work,very essence of life,thanks for sharing this,I never saw this palindromic principle demonstrated like this before
At the point at which you’re looking at x^4 + 6x^3 + 11x^2 + 6x + 1, I would be tempted to “try” (x^2 + cx + 1)^2 and solve for c if it looks promising. We expect a square, by the way the question is posed. Also, all coefficients are positive, so a little thought tells us we can’t have any negative coefficients in the factor, so we don’t have to try -1 (or -x^2). This gives us x^4 + 2cx^3 + (c^2 + 2)x^2 + 2cx + 1, immediately giving 2c=6, c=3, which thankfully is compatible with c^2+2=11. So the guess is right, and we just plug 96 into x^2+3x+1
Very good 👍❤
I did it an unconventional way (and I don't know if it's "legal" in maths) :
I've tried with x=1, and the result was 25, 5², or (1*4 +1)²
Then I tried with x=2, and the result was 121, 11² or (2*5 +1)²
So I theorized the result could be (x*(x+3) + 1)² and I wrote the calcul :
x(x+1)(x+2)(x+3) + 1 = (x(x+3) + 1)^2
I developped both sides and got
x^4 + 6x^3 + 11x^2 + 6x + 1 = x^4 + 6x^3 + 11x^2 + 6x + 1
The same calcul on both sides, so my assumption seems correct.
Put under the square root, it gives x(x+3) + 1.
Replace x with 96, I have 96*99 + 1, or 9504 + 1 or 9505.
Edit : minor corrections
You'll need to show that x(x+d)(x+2d)(x+3d) +d⁴ is a perfect square
I'm so happy I found this channel
U are Awesome I like your unique problem solving methods ❤
I gave this one a try and set x to be the second number of the four, because (x-1)(x+1) was a nice, easy product to deal with. The factoring was a bit guess-and-check, but since it started with x^4 and ended with 1, the square root was obviously x^2+ax(+/-)1, and the coefficients I had were all ones or twos, so I only had four meaningful options for a. The end result was x^2+x-1, which looks like a nice, neat answer.
I wish I had you as teacher in University classes, for real.
Amazing!
Bless you
My way for this state is:
Suppose those numbers ( 99,96,97,98 )that are 100
So sq(100^4) is 100^2=10000
But we know those numbers are not really 100 ,they less &soon
Then the answer is less&soon 10,000
I liked your approach but I was tempted by how close the numbers were to 100
this is my attack on this problem
sqrt(96*97*98*99 + 1)
this is actually a bit simpler if you take h = 100 and substitute:
sqrt((h-4)(h-3)(h-2)(h-1) + 1)
then switch the pairs to make pairs with numbers that sum up to 5
sqrt( (h-4)(h-1) * (h-3)(h-2) + 1 )
then multiply the pairs out
sqrt( (h^2 - 5h + 4)(h^2 - 5h + 6) + 1 )
then we notice the first two terms of each polynomial in the product are the same (h^2 - 5h)
so substitute n = h^2 - 5h which substituting back 100 for h works out to
n = h^2 - 5h = (100^2) - (5*100) = 10000 - 500 = 9500
so actually n = 9500
putting n in this expression works out to
sqrt((n+4)(n+6) + 1)
multiply the binomials out
sqrt(n^2 + 10n + 24 + 1)
adding the two constants
sqrt(n^2 + 10n + 25)
now we see that this is equivalent to
sqrt(n^2 + 2 * 5 * n + 5^2)
so we actually have a square of a binomial
sqrt( (n + 5) ^ 2 )
the square and the square root cancel out
n + 5
substitute back n = 9500 which we figured out when getting n from h^2 - 5h
9500 + 5
which makes the answer
9505
That's sooooo cool. Like wtf, algebra , how can you do this
Graphs of the function we start with and of the function we end up with are different.
😳 woah!
fav teacher
I was thinking of the product of sums for weighted choice of x at some middle coefficient to let x = 97 or x = 98 but that fails for even multiplied x sums. So this method turns out better. Plus when I divided out the x^2 to get an answer I forgot to remuktiply it back in. I was doing the solution in R^2 and then the multiplied 11 coefficient dividing out by x^2 I just ignored that in finding the final results I saw you get with R^2 = x^2 multiplied back in. I need this brainwork practicing because I do mathematics statements errors like that too often! 😬
So it's effectively 97² + 96 in the end
If you take the absolute value, you get a valid expression of R(x)= | x²+3x+1 | = √(x.(x+1).(x+2).(x+3)+1) for any positive, null or negative real x. The square root is always defined since the palindromic polynômial x⁴+6x³+11x²+6x+1 is never negative whatever the value of x. But the absolute value is mandatory in R(x) expression since polynômial x²+3x+1 is négative when x is in ] (-3-√5)/2 ; (-3+√5)/2 [.
Note that R(x)=1 for x in { -3 -2 -1 0 } only.
Furthermore R(x)=0 at x=(-3±√5)/2 respectively x ≈ -2.6180 and x ≈ -0.38197
next interesting problem is to found how positive and negative integers p>0 and n
Very good
Thanks
You are Cool
Maybe x=97 is easier.
But everybody is a sinner!
The formula works for all strictly positive integer.
Does it works for 0 ? well, strangely enough yes, although, the proof doesn't work if x = 0
What about negative numbers ?
for -3, -2, -1 and 0, R is 1, no need for a formula
for any number below -3, we are multiplying 4 negative numbers, which is the same as multiplying 4 positive numbers, but |x| is now the biggest of the 4
so with y = -(x+3)
R = y^2 + 3y +1
or
R = (x+3)^2 - 3x -8
10:30 I swear I won't consent to sinners' proposals :D
If a,b,c and d are in arithmetic progression with common difference k, then abcd+k⁴ is a perfect square.
What theorem or lemma is this?
@@PrimeNewtons I am not sure about it. I had a discussion about this problem with my friend and we just discovered it.
I'm sure one can do a proof for it
@@PrimeNewtonsYa I have already tried out the proof. It works!
@@PrimeNewtons Ya I have alreaday tried out the proof. It works!
Palindromic Property is a little complicated to me , so i try this one , lucky , it works.
(x-d)xy(y+d)+d^4 where x+d=y
=(x^2-xd)(y^2+yd)+d^4
=(xy)^2+x^2yd-xy^2d-xyd^2+d^4
=(xy)^2-xyd(y-x)-xyd^2+d^4
=(xy)^2-2xyd^2+d^4
=(xy-d^2)^2