[dark_magician_92]

say 'n' is an integer. then can n + 3 and n^2 + 3 both be perfect cubes?

use spoilers.

"empty here"

just kidding

[dark_magician_92]

and one more thing, give reasoning for your answers!!

[dark_magician_92]

and hints are also available when u ask.

[wolfgang]

n= 1

[dark_magician_92]

perfect cube not perfect square wolfgang , and reasoning also needed. now i am off to sleep. see ya tmrw

[k-man]

It's not possible. Here is the explanation

Suppose there is an integer x such that x³ = n+3. Then (x³-3) = n.

We need to determine if it's possible that (x³-3)²+3 is a perfect cube. Expanding the expression we get x⁶-6x³+12

The highest degree of x in this expression is 6, so if this expression is a perfect cube then it can be expressed in the form (ax²+bx+c)³, where a, b and c are some constants. If we can find such a, b and c that will transform (ax²+bx+c)³ into x⁶-6x³+12 then we have the solution.

Expanding (ax²+bx+c)³ we get a³x⁶+3a²bx⁵+3a²cx⁴+3ab²x⁴+6abcx³+3ac²x²+b³x³+3b²cx²+3bc²x+c³

The coefficient for x⁶ is 1, so a=1. The coefficient for x⁵ is 0, so b must be 0. But if b=0 then the expression is simplified to x⁶+3cx⁴+3ac²x²+c³ and cannot be transformed into x⁶-6x³+12 anymore because the coefficient for x³ is 0. Also, we need to get rid of x⁴ and x², so c must be 0 too.

Basically, there are no such a, b and c that would allow to transform (ax²+bx+c)³ into x⁶-6x³+12 and therefore x⁶-6x³+12 cannot be a perfect cube.

[k-man]

I found a flaw in my reasoning, so I will have to think about it some more later...

[Guest]

Let x be the cube root of n+3 and y be the cube root of n^2+3.

I used spreadsheet to solve for n and y for all x from -50 to 50. There were no integral y solutions. There is an interesting pattern. For all absolute x greater than 6, y = x^2-2/x. For x between -6 and 6, this approximation is less accurate but is pretty exact for all other x. And for any x greater than 2, the term 2/x will always be less than 1 so for integral x, y will always have a fraction or decimal. As x approaches infinity, this decimal approaches zero and y approaches x^2. y=x^2 is what would be the case if x were simply cube root of n and y cube root of n^2 as well, the addition of the 3 in both equations just throws it off a bit, but for larger and larger n that matters less and less and y gets closer and closer to x^2, but never quite reaches it, thus it will always be a non-integer.

Edited July 3, 2011 by Nana7

[Guest]

By definition of a perfect cube, (n² + 3) and (n + 3) must be integers, and so then must be their difference. The difference between (n² + 3) - (n + 3) is n(n - 1).

The factors of n(n-1) are n and n-1. The only primes that have a difference of 1 is 3 and 2, thus for n to be an integer in the expression n(n-1), n must equal 3. Neither [3]+3 = 6 nor [3]^2+3 = 12 are perfect cubes.

Ergo, both (n² + 3) and (n + 3) can not be perfect cubes.

[dark_magician_92]

By definition of a perfect cube, (n² + 3) and (n + 3) must be integers, and so then must be their difference. The difference between (n² + 3) - (n + 3) is n(n - 1).

The factors of n(n-1) are n and n-1. The only primes that have a difference of 1 is 3 and 2, thus for n to be an integer in the expression n(n-1), n must equal 3. Neither [3]+3 = 6 nor [3]^2+3 = 12 are perfect cubes.

Ergo, both (n² + 3) and (n + 3) can not be perfect cubes.

is it really true that the difference between 2 perfect squares equal to the product of 2 prime nos??

[dark_magician_92]

Let x be the cube root of n+3 and y be the cube root of n^2+3.

I used spreadsheet to solve for n and y for all x from -50 to 50. There were no integral y solutions. There is an interesting pattern. For all absolute x greater than 6, y = x^2-2/x. For x between -6 and 6, this approximation is less accurate but is pretty exact for all other x. And for any x greater than 2, the term 2/x will always be less than 1 so for integral x, y will always have a fraction or decimal. As x approaches infinity, this decimal approaches zero and y approaches x^2. y=x^2 is what would be the case if x were simply cube root of n and y cube root of n^2 as well, the addition of the 3 in both equations just throws it off a bit, but for larger and larger n that matters less and less and y gets closer and closer to x^2, but never quite reaches it, thus it will always be a non-integer.

yes u r right, but i asked for a mathematical reasoning, not using spreadsheets, no offence though

Edited July 3, 2011 by dark_magician_92

[Guest]

n+3 is a red herring! n^2+3 can never be a perfect cube.

[dark_magician_92]

n+3 is a red herring! n^2+3 can never be a perfect cube.

wherez d proof? smart though!

[Guest]

There are two definitions to a perfect cube (and perfect square). First is where the cube (or respectively, square) root must be an integer, and second where it does not need be an integer, but must be a rational number. I shall assume here the more limited definition that a perfect cube must be an integer.

(n² + 3) - (n + 3) = (n - 1)n

The difference between two cubes can be expressed as:

a³ - b³ = (a - b)(a + b)²

Let n = (a + b)², then by subsitution:

(a - b) = (a + b)² - 1

The equation can be formed as the quadratic:

a² + (2b - 1)a + (b² + b - 1) = 0.

The discriminant Δ of the quadratic is -8b + 5.

From the nature of the discriminant, for the polynomial to have at least one real root, i.e. Δ >= 0,

b <= 5/8.

In order for b³ to be a perfect cube _{(see definition above)}, b must be an integer < 5/8, and in order for a to be an integer

-8b + 5 must be a perfect square of an integer: x² = -8b + 5.

Let b = kx to form the quadratic equation: x^2 + 8kx - 5 = 0.

(x + i√5)² = x² + 2i√5 - 5, thus, in order for the quadratic equation to be a perfect square 8k = 2i√5, i.e., k = i√5/4. Substituting k = i√5/4 into the quadratic gives: x² + 2i√5x - 5 = 0. Solving for the roots of x, gives x = -i√5. Substituting this value back into x² = -8b + 5 gives b = 5/4, which is neither an integer nor would b be < 5/8. Ergo both (n² + 3) and (n + 3) can not be perfect cubes.

Edited July 3, 2011 by Dej Mar

[wolfgang]

If n=5

so..

5+3 = 8 which is a perfect cube

5^2 +3= 28

28= 27+1( and both are perfect cubes)

Edited July 3, 2011 by wolfgang

[dark_magician_92]

There are two definitions to a perfect cube (and perfect square). First is where the cube (or respectively, square) root must be an integer, and second where it does not need be an integer, but must be a rational number. I shall assume here the more limited definition that a perfect cube must be an integer.

(n² + 3) - (n + 3) = (n - 1)n

The difference between two cubes can be expressed as:

a³ - b³ = (a - b)(a + b)²

Let n = (a + b)², then by subsitution:

(a - b) = (a + b)² - 1

The equation can be formed as the quadratic:

a² + (2b - 1)a + (b² + b - 1) = 0.

The discriminant Δ of the quadratic is -8b + 5.

From the nature of the discriminant, for the polynomial to have at least one real root, i.e. Δ >= 0,

b <= 5/8.

In order for b³ to be a perfect cube _{(see definition above)}, b must be an integer < 5/8, and in order for a to be an integer

-8b + 5 must be a perfect square of an integer: x² = -8b + 5.

Let b = kx to form the quadratic equation: x^2 + 8kx - 5 = 0.

(x + i√5)² = x² + 2i√5 - 5, thus, in order for the quadratic equation to be a perfect square 8k = 2i√5, i.e., k = i√5/4. Substituting k = i√5/4 into the quadratic gives: x² + 2i√5x - 5 = 0. Solving for the roots of x, gives x = -i√5. Substituting this value back into x² = -8b + 5 gives b = 5/4, which is neither an integer nor would b be < 5/8. Ergo both (n² + 3) and (n + 3) can not be perfect cubes.

sorry for me being lazy to not read the complete soln but

a³ - b³ = (a - b)(a ²+ b² +ab)

and not what you wrote

[Guest]

sorry for me being lazy to not read the complete soln but

a³ - b³ = (a - b)(a ²+ b² +ab)

and not what you wrote

(n² + 3) - (n + 3) = (n - 1)n

The difference between two cubes can be expressed as: a³ – b³ = (a - b)(a² + ab + b²)

Let the factor n be equal to the factor (a² + ab + b²), then by substitution: (a - b) = a² + ab + b² – 1

The equation can be formed as a quadratic in terms of the indeterminant a: a² + (b – 1)a + (b² + b – 1) = 0

The discriminant of the quadratic for the indeterminant a is the quadratic -3b² - 6b + 5 in terms of the indeterminant b. The discriminant for the latter quadratic is 96, which by the nature of the discriminant indicates that the quadratic for -3b² - 6b + 5 = 0 has two real roots, which can be found to be b₁ = -1 + 2/3•√6 and b₂ = -1 - 2/3•√6. The root b₂ = -1 - 2/3•√6 provides no real root for a, thus the remaining root b₁ = -1 + 2/3•√6 is examined: a² + (2/3•√6 - 2)a + (–2/3•√6 + 5/3) = 0. The discriminant for this quadratic is 0 indicating the polynomial has only one real root: (1 - 1/3•√6) which is neither an integer or rational,

ergo both (n² + 3) and (n + 3) can not be perfect cubes.

[dark_magician_92]

this works!!

however another go:-

If n + 3 and n2 + 3 are both perfect cubes then their product must also be a perfect cube.

So consider (n + 3)(n² + 3) = n³ + 3n² + 3n + 9 = (n + 1)³3 + 8.

This can be a perfect cube only if it is 8 more than another perfect cube, namely (n + 1)3.

The only pairs of perfect cubes that differ by 8 are (−8, 0) and (0, 8).

So we must have

(n + 1)³ = (−2)³ n = −3; or

(n + 1)³ = 0³ n = −1.

For neither of these solutions is n2 + 3 a perfect cube.

Therefore, if n is an integer, n + 3 and n2 + 3 cannot both be perfect cubes.

[Guest]

(n² + 3) - (n + 3) = (n - 1)n

The difference between two cubes can be expressed as: a³ – b³ = (a - b)(a² + ab + b²)

Let the factor n be equal to the factor (a² + ab + b²), then by substitution: (a - b) = a² + ab + b² – 1

The equation can be formed as a quadratic in terms of the indeterminant a: a² + (b – 1)a + (b² + b – 1) = 0

The discriminant of the quadratic for the indeterminant a is the quadratic -3b² - 6b + 5 in terms of the indeterminant b. The discriminant for the latter quadratic is 96, which by the nature of the discriminant indicates that the quadratic for -3b² - 6b + 5 = 0 has two real roots, which can be found to be b₁ = -1 + 2/3•√6 and b₂ = -1 - 2/3•√6. The root b₂ = -1 - 2/3•√6 provides no real root for a, thus the remaining root b₁ = -1 + 2/3•√6 is examined: a² + (2/3•√6 - 2)a + (–2/3•√6 + 5/3) = 0. The discriminant for this quadratic is 0 indicating the polynomial has only one real root: (1 - 1/3•√6) which is neither an integer or rational,

ergo both (n² + 3) and (n + 3) can not be perfect cubes.

You my dismiss this, my previous solution. I see where I made an error. (Saw it immediatly upon posting, but was too tired due to having to take care of my 3 year old daughter that I did not have the energy to bother removing.)

[Guest]

The digital root of a cube has a periodicity of 3 with the cycle (1, 8, 9).

The digital root of a number of the form x(n + 3), where x and n are integers, has a periodicity of 9 with the cycle

(4, 5, 6, 7, 8, 9, 1, 2, 3).

(1, 8, 9, 1, 8, 9, 1, 8, 9) digital roots of a cube

All three digital roots of a cube correspond in the cycle, thus any cube may be represented in the form x(n + 3) (Not all x(n + 3) are cubes!).

The digital root of a number of the form y(n² + 3), where y and n are integers, has a periodicity of 9 with the cycle of

(4, 7, 3, 1, 1, 3, 7, 4, 3).

(1, 8, 9, 1, 8, 9, 1, 8, 9) digital roots of a cube

Therefore, the digital root of a cube of the form y(n² + 3) must have a digital root of 1.

(4, 5, 6, 7, 8, 9, 1, 2, 3) digital roots of x(n + 3)

(4, 7, 3, 1, 1, 3, 7, 4, 3) digital roots of y(n² + 3)

(1, 8, 9, 1, 8, 9, 1, 8, 9) digital roots of a cube

x(n+3) and y(n² + 3) do not both share any of the same digital roots corresponding to the periodicity cycle of 9 with a cube, therefore both (n² + 3) and (n + 3) can not be perfect cubes.

In addition, if (n² + 3) and (n + 3) are both perfect cubes, then so must (n² + 3)×(n + 3) which equals (n+1)³ + 8. But as (n+1)³ is a cube of periodicity 3, the offset, + 8, modulo 3 would need be 0 in order for (n+1)³ + 8 to be a perfect cube as well, but it could not be, thus both (n² + 3) and (n + 3) can not be perfect cubes.

Edited July 5, 2011 by Dej Mar

[Guest]

The digital root of a cube has a periodicity of 3 with the cycle (1, 8, 9).

The digital root of a number of the form x(n + 3), where x and n are integers, has a periodicity of 9 with the cycle

(4, 5, 6, 7, 8, 9, 1, 2, 3).

(1, 8, 9, 1, 8, 9, 1, 8, 9) digital roots of a cube

All three digital roots of a cube correspond in the cycle, thus any cube may be represented in the form x(n + 3) (Not all x(n + 3) are cubes!).

The digital root of a number of the form y(n² + 3), where y and n are integers, has a periodicity of 9 with the cycle of

(4, 7, 3, 1, 1, 3, 7, 4, 3).

(1, 8, 9, 1, 8, 9, 1, 8, 9) digital roots of a cube

Therefore, the digital root of a cube of the form y(n² + 3) must have a digital root of 1.

(4, 5, 6, 7, 8, 9, 1, 2, 3) digital roots of x(n + 3)

(4, 7, 3, 1, 1, 3, 7, 4, 3) digital roots of y(n² + 3)

(1, 8, 9, 1, 8, 9, 1, 8, 9) digital roots of a cube

x(n+3) and y(n² + 3) do not both share any of the same digital roots corresponding to the periodicity cycle of 9 with a cube, therefore both (n² + 3) and (n + 3) can not be perfect cubes.

In addition, if (n² + 3) and (n + 3) are both perfect cubes, then so must (n² + 3)×(n + 3) which equals (n+1)³ + 8. But as (n+1)³ is a cube of periodicity 3, the offset, + 8, modulo 3 would need be 0 in order for (n+1)³ + 8 to be a perfect cube as well, but it could not be, thus both (n² + 3) and (n + 3) can not be perfect cubes.

That the integers x & y are unnecessary, especially since it is only true that "thus any cube may be represented in the form x(n + 3)" if x can equal 1. Further, the problem was proving that they both could not be perfect cubes, your proof only showed that the could not have the same cube. I like your approach in the last two lines better but your conclusion seems wrong here because, if (n+1)³ can have a digital root of 1. Adding 8, yields digital root of 9 which does present a possible, though not necessarily, a perfect cube. However, since (n+1) must be equal to or greater than 2, the difference between (n+1)³ and the next perfect cubes must be greater than 8

Unless I am missing something.

Edited July 5, 2011 by thoughtfulfellow

[Guest]

I guess my sleep was not so decent.

A final attempt...

If (n² + 3) and (n + 3) are both perfect cubes, then so must (n² + 3)×(n + 3) be a perfect cube.

The product (n² + 3)×(n + 3) can also be expressed as the sum of two cubes (n + 1)³ + 2³.

Fermat's Last Theorem states that no nontrivial integer solution exists for the equation aⁿ + bⁿ = cⁿ where n is an integer greater than 2. The theorem was proved by Andrew Wiles in the mid-1990s, ergo, unless Wiles proof is invalid, (n²+ 3) and (n + 3) can not be both perfect cubes.

Edited July 5, 2011 by Dej Mar

[superprismatic]

I guess my sleep was not so decent.

A final attempt...

If (n² + 3) and (n + 3) are both perfect cubes, then so must (n² + 3)×(n + 3) be a perfect cube.

The product (n² + 3)×(n + 3) can also be expressed as the sum of two cubes (n + 1)³ + 2³.

Fermat's Last Theorem states that no nontrivial integer solution exists for the equation aⁿ + bⁿ = cⁿ where n is an integer greater than 2. The theorem was proved by Andrew Wiles in the mid-1990s, ergo, unless Wiles proof is invalid, (n²+ 3) and (n + 3) can not be both perfect cubes.

This is a very nice proof! As Paul Erdős would have said, "This proof is in The Book!".

[Guest]

I guess my sleep was not so decent.

A final attempt...

If (n² + 3) and (n + 3) are both perfect cubes, then so must (n² + 3)×(n + 3) be a perfect cube.

The product (n² + 3)×(n + 3) can also be expressed as the sum of two cubes (n + 1)³ + 2³.

Fermat's Last Theorem states that no nontrivial integer solution exists for the equation aⁿ + bⁿ = cⁿ where n is an integer greater than 2. The theorem was proved by Andrew Wiles in the mid-1990s, ergo, unless Wiles proof is invalid, (n²+ 3) and (n + 3) can not be both perfect cubes.

I agree with superprismatic! This solution looks flawless. It's also concise and clear. Congratulations!

[dark_magician_92]

best solution so far, well done