MHB Prove $m^5+3m^4n-5m^3n^2-15m^2n^3+4mn^4+12n^5$ ≠ 33

  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    Algebra Challenge
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Prove that $m^5+3m^4n-5m^3n^2-15m^2n^3+4mn^4+12n^5$ is never equal to 33.
 
Mathematics news on Phys.org
anemone said:
Prove that $m^5+3m^4n-5m^3n^2-15m^2n^3+4mn^4+12n^5$ is never equal to 33.

First let us factor $m^5+3m^4n-5m^3n^2-15m^2n^3+4mn^4+12n^5$
$m^5+3m^4n-5m^3n^2-15m^2n^3+4mn^4+12n^5$
= $m^4(m+3n) - 5m^2n^2(m+3n) + 4n^4(m+3n)$
= $(m+3n)(m^4-5m^2n^2+4n^4)$
= $(m+3n)(m^2-4n^2)(m^2-n^2)$
= $(m+3n)(m-2n)(m+2n)(m-n)(m+n)$
= $(m-2n)(m-n)(m+n)(m+2n)(m+3n)$
for n = 0 the value is $m^5$ and 33 is not a $5^{th}$ power
for n not zero above 5 values are differnt and 33 is product of atmost 4 mumbers $( (- 1) * 1 * (-3) * 11$ and hence product
cannot be 33
 
kaliprasad said:
First let us factor $m^5+3m^4n-5m^3n^2-15m^2n^3+4mn^4+12n^5$
$m^5+3m^4n-5m^3n^2-15m^2n^3+4mn^4+12n^5$
= $m^4(m+3n) - 5m^2n^2(m+3n) + 4n^4(m+3n)$
= $(m+3n)(m^4-5m^2n^2+4n^4)$
= $(m+3n)(m^2-4n^2)(m^2-n^2)$
= $(m+3n)(m-2n)(m+2n)(m-n)(m+n)$
= $(m-2n)(m-n)(m+n)(m+2n)(m+3n)$
for n = 0 the value is $m^5$ and 33 is not a $5^{th}$ power
for n not zero above 5 values are differnt and 33 is product of atmost 4 mumbers $( (- 1) * 1 * (-3) * 11$ and hence product
cannot be 33

Very well done, kaliprasad!(Cool)
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top