let x,y,z be cube roots of three distinct prime integers.?

Question

Show that x,y,z are never three terms (not necessarily consecutive)of an A.P.

Answer 1

Suppose, without loss of generality, that x < y < z. Let c denote the cube root function. Then, there are primes p1 < p2 < p3 such that x = c(p1), y = c(p2), z = c (p3). Suppose, x y and z are consecutive terms of an AP. Then,

2y = x + z (1) and, therefore,
2c(p2) = c(p1) + c(p3) (2)

If we cube both members of (1), we get

8 y^3 = x^3 + 3x^2 z + 3x z^2 +zy^3, and, therefore,
8p2 = p1 + 3 c(p1^2 p3) + 3c(p1 p3^2) + p3,so that
8p2 - p1 - p3 = 3c(p1 p3) [c(p1) + c(p3)]

In virtue of (2), it follows that

8p2 - p1 - p3 = 3c(p1 p3) [2 c(p2)]
8p2 - p1 - p3 = 6 c(p1 p2 p3)

In the left side we have an integer. Since p1, p2, p3 are primes p1 * p2 * p3 is not a perfect cube, which implies c(p1 p2 p3) is irrational and so is the right hand side. Hence, we have reached a contradiction that proves x, y and z cannot be consecutive terms of an AP.

Actually, it's not necessary that p1, p2 and p3 be primes. It suffices that p1 * p2 * p3 is not a perfect cube.

This proof is similar to Duke's one.

Answer 2

To "Show that x,y,z are never three terms (not necessarily consecutive)of an A.P" I would say it is necessary and sufficient to show that (z-y)/(y-x) is irrational. If it is some rational number a/b, the for some real number w, x + bw = y and y + aw = z, so x,y,z would be in an arithmetic progression with common difference w.

Hmmm... I am not sure how to use that fact above, but I will think some more about this. Thanks for the neat problem!

Very nice work by the two folks below me! So there is no need for me to continue. Of the two Duke's is more complete I think, though I think the statement "If P, Q, R are terms, then one of the differences Q-P and R-Q is an integer multiple of the other " should have integer replaced with rational, but it looks like everything after that still works.

Answer 3

The approach in both answers above is correct and fruitful and Lobosito has almost reached the answer, his final step only missing. I'll use the known fact that a cubic root of an integer-non-perfect cube is irrational. Let
p < q < r are primes and let for convenience
P = cube_root(p);
Q = cube_root(q);
R = cube_root(r), of course P < Q < R also.
If P, Q, R are terms, then one of the differences Q-P and R-Q is an ***integer*** (CORRECTION - rational) multiple of the other as both answers above say, let
R - Q = n(Q - P), here n - ***natural*** (CORRECTION - rational). Then
(!!!) nP + R = (n + 1)Q, raising to the 3rd degree we obtain:
n³ p + 3nPR(nP + R) = (n + 1)³ r, or
3nPR(nP + R) = (n + 1)³ r - n³ p, here the right side obviously is an ***integer*** (CORRECTION - rational), so
PR(nP + R) = ((n + 1)³ r - n³p) / (3n), or, using (!!!)
PQR(n + 1) = ((n + 1)³ r - n³p) / (3n), or, finally:
PQR = ((n + 1)³ r - n³p) / (3n(n + 1)).
Here is the required contradiction: the right side is rational number, the left - cubic root of p*q*r, which IS NOT A PERFECT CUBE /p,q,r - distinct!/, i.e. irrational.

P.S. (EDIT) Phineas Bogg: Thanks for the correction, You are quite right and gave me a chance to make the proof faultless!
Author: Thanks for the neat problem, I forgot the original notations x,y,z and used P,Q,R instead.

P.S.(2) Steiner: You consider only consecutive terms, but the problem requires to prove the statement for ANY 3 TERMS (read the condition carefully) - the same principle works with slight modifications.

Answer 4

Hmmm... I am not sure how to use that fact above, but I will think some more about this. Thanks for the neat problem!

simply, suppose (z-y)/(y-x) = s, s different than 0
so z-y=s(y-x)
z+sx = y(s+1)
(z+sx)^3 = y^3(s+1)^3
z^3+s^3x^3+3z^2sx+3(sx)^2z = y^3(s+1)^3

z^2x+sx^2z = (y^3(s+1)^3-z^3-s^3x^3)/3s

The term in right hand is in Q, so the term in left hand must be in Q.
Let a = z^2x and b=sx^2z.
Then a+b and ab are in Q, so they are roots of the polynomial
u^2-(a+b)u+ab=0, with coefficients in Q.
But a is the root of u^3-a^3=0 with coefficients in Q, which by rational root theorem is irreducible.

Answer 5

construction on gianlino and MathPHD. by utilizing Wilson's theorem (p-a million)!=-a million mod p for all p and so (p-2)!=a million mod p (Eq1) for all p. So evaluate n=p-2, with p primes in worry-free words. If the statement were pretend, then (p-2)!-a million=p^e by utilizing Eq1 for all primes better than some volume N. Now evaluate both aspect mod 4, then p^e= -a million mod 4 for all primes better than N. p^2 = a million mod 4 provided that p is an peculiar best, so e might want to be peculiar, and all primes better than N might want to be congruent to -a million mod 4. reported otherwise, the type of primes congruent to a million mod 4 might want to be finite. yet there are infinitely many primes = a million mod 4 (and also you grants a thoroughly worry-free evidence without Dirichlet). So, there is not any volume N such that (p-2)!-a million=p^e for all primes p >N, ie, there might want to be infinitely many primes p so as that (p-2)!-a million is divisible by utilizing another best q as well as p. we are performed.