Number Theory: Difference of Perfect Squares

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  • #1
S.R
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Homework Statement

:[/B]
Determine whether there exists an integer x such that x^2 + 10 is a perfect square.

Homework Equations

:[/B]
N/A

The Attempt at a Solution

:[/B]
Assume x^2 + 10 = k^2 (a perfect square).

Solve for x in terms of k:
x = ±sqrt(k^2 - 10)

Since k is an integer and k^2 - 10 > 0, k > sqrt(10) > 3.

From here (and multiple other approaches), I'm not sure how to continue. Any help in proving whether such an integer value of x exists would be appreciated.
 

Answers and Replies

  • #2
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Hmmm....I've got something in mind ,
You have :
##k^2=x^2+10##
And thus ,
##k^2-x^2=10##
##(k-x)(k+x)=10##
And since 10=5×2 ,
Therefore , we can say that
Either ##k-x=2 ## and ##k+x=5##
Or ##k+x=2## and ##k-x=5##
Which will give you fractional values of x and k , which don't satisfy the condition .
Therefore no integer should satisfy this equation.
 
  • #3
S.R
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Ah, ok. Thank-you.
 
Last edited:
  • #4
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Hmmm....I've got something in mind ,
You have :
##k^2=x^2+10##
And thus ,
##k^2-x^2=10##
##(k-x)(k+x)=10##
And since 10=5×2 ,
Therefore , we can say that
Either ##k-x=2 ## and ##k+x=5##
Or ##k+x=2## and ##k-x=5##
No, this doesn't follow. If a * b = 0, then you can definitely say something about a being 0 or b being 0, but if a * b = m, with m ≠ 0, then you effectively don't know anything about a and b.
mooncrater said:
Which will give you fractional values of x and k , which don't satisfy the condition .
Therefore no integer should satisfy this equation.
 
  • #5
33,721
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Homework Statement

:[/B]
Determine whether there exists an integer x such that x^2 + 10 is a perfect square.

Homework Equations

:[/B]
N/A

The Attempt at a Solution

:[/B]
Assume x^2 + 10 = k^2 (a perfect square).
So k2 - x2 = 10, where both k and x are integers.

Since k is an integer, we can write it as k = x + a, for some integer a. What can you say about ##k^2 - x^2 = (x + a)^2 - x^2##? Can you show that this difference can never be 10?


S.R said:
Solve for x in terms of k:
x = ±sqrt(k^2 - 10)

Since k is an integer and k^2 - 10 > 0, k > sqrt(10) > 3.

From here (and multiple other approaches), I'm not sure how to continue. Any help in proving whether such an integer value of x exists would be appreciated.
 
  • #6
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Does k have to be an integer or can k be rational? For n>=5 the difference between n squared and n+1 squared is 2n+1>10.
 
  • #7
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Does k have to be an integer or can k be rational? For n>=5 the difference between n squared and n+1 squared is 2n+1>10.
I believe that "perfect square" as used in the OP means the square of an integer.
 
  • #8
S.R
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I attempted this approach, but I guess I'll try again here.

(x+a)^2 - x^2 = 10
x^2 + (2a)x + a^2 - x^2 = 10
(2a)x + a^2 = 10
a(2x + a) = 10

From here, I can presumably set a equal to factors of 10, right?
 
  • #9
haruspex
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No, this doesn't follow. If a * b = 0, then you can definitely say something about a being 0 or b being 0, but if a * b = m, with m ≠ 0, then you effectively don't know anything about a and b.
I'm puzzled by your response here. The only thing I see wrong with mooncrater's approach is that it overlooked another possible factorisation. Isn't your x+a method effectively the same?
S.R, there are certain facts about integers that are handy to remember. Perfect squares only take a few values modulo 8. What are they?
 
  • #10
S.R
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I'm not sure what you mean by "perfect squares only take a few values modulo 8"?
 
  • #11
haruspex
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I'm not sure what you mean by "perfect squares only take a few values modulo 8"?
For now, just consider modulo 4, since it is adequate for this problem.
Take an even number, 2n, and square it. What is the result modulo 4? (You understand what is meant by modulo, yes?). Do the same for 2n+1.
 
  • #12
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I'm puzzled by your response here. The only thing I see wrong with mooncrater's approach is that it overlooked another possible factorisation.
My main objection is that if a * b = 0, you definitely know something about a or b. OTOH, if a * b = N, there are an infinite number of pairs of numbers that work. In that post we had a * b = 10, from which mooncrater concluded that the two numbers have to be 5 and 2. Of course -5 and -2 work, as do 10 and 1, -10 and -1, 1/2 and 20, and many, many more. If he had included the other pair of positive integers, and mentioned that he was limiting the search to positive integers only, I don't think I would have said anything.
haruspex said:
Isn't your x+a method effectively the same?
 
  • #13
S.R
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(2n)^2 = 4n^2
(4n^2) mod 4 = 0 (assuming n is an integer).

(2n+1)^2 = 4n^2 + 4n + 1.
(4n^2 + 4n + 1) mod 4 = 1 (assuming n is an integer).

Not particularly sure how this fact helps, though?

EDIT: Oh, if we assume k is a perfect square, then k mod 4 = 0 or k mod 4 = 1 (depending on whether k is even or odd).

1) Assume x is even
Let x = 2n, where n is an integer
x^2 + 10 = 4n^2 + 10

(4n^2 + 10) mod 4 = 2 ≠ 0 and thus, x^2 + 10 is not a perfect square if x is even.

2) Assume x is odd
Let x = 2n+1, where n is an integer
x^2 + 10 = 4n^2 + 4n + 11

(4n^2 + 4n + 11) mod 4 = 3 ≠ 1 and thus, x^2 + 10 is not a perfect square if x is odd.

Since x^2 + 10 is not a perfect square if x is even nor odd, we can conclude that no such integer x exists such that x^2 + 10 is a perfect square.

Is this correct?
 
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  • #14
haruspex
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(2n)^2 = 4n^2
(4n^2) mod 4 = 0 (assuming n is an integer).

(2n+1)^2 = 4n^2 + 4n + 1.
(4n^2 + 4n + 1) mod 4 = 1 (assuming n is an integer).

Not particularly sure how this fact helps, though?

EDIT: Oh, if we assume k is a perfect square, then k mod 4 = 0 or k mod 4 = 1 (depending on whether k is even or odd).

1) Assume x is even
Let x = 2n, where n is an integer
x^2 + 10 = 4n^2 + 10

(4n^2 + 10) mod 4 = 2 ≠ 0 and thus, x^2 + 10 is not a perfect square if x is even.

2) Assume x is odd
Let x = 2n+1, where n is an integer
x^2 + 10 = 4n^2 + 4n + 11

(4n^2 + 4n + 11) mod 4 = 3 ≠ 1 and thus, x^2 + 10 is not a perfect square if x is odd either.

Since x^2 + 10 is not a perfect square if x is even nor odd, we can conclude that no such integer x exists such that x^2 + 10 is a perfect square.

Is this correct?
Yes.
 
  • #15
S.R
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Thank-you for the help!
 

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