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- Thread starter scox
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AKG

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The number we're looking for, let's call it [itex]a[/itex]. Let's call the perfect square [itex]b^2[/itex]. So:scox said:

[tex]a^2 \pm 5 = b^2[/tex]

[tex]a^2 - b^2 = \mp 5[/tex]

[tex](a - b)(a + b) = \mp 5[/tex]

Now, if a > b, then a - b > 0, and a + b > 0, therefore (a - b)(a + b) > 0, and thus is not equal to -5, so it's +5. If a < b, then it's -5. We can basically see that whether we choose a > b or b > a, it's not really going to matter. In other words, if we chose a > b, we would find a certain value for a and a certain value for b. If we chose b > a, we would find the same values, but just swapped around, but it really doesn't matter which one we call a and which one we call b. So, let's just go with a > b:

[tex](a - b)(a + b) = 5[/tex]

Now, 5 is a prime number, and the only two numbers that divide 5 are 1 and 5, so we can conclude a - b = 1, and a + b = 5. Solve, and we get a = 3, b = 2. As you can see, [itex]3^2 - 5 = 2^2[/tex]. If we had gone the other way, and gotten a = 2, b = 3, we would simply have [itex]2^2 + 5 = 3^2[/itex] and since the question said that we can either add or subtract 5, either answer should be acceptable.

- #3

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1, 2, 3, 4, 5, 6, 7

1, 4, 9, 16, 25, 36, 49

that doesnt help much except to see that there isnt a square that is +/- 10 of another square. i dont see how any number can into this definition, (x*x)+-5=c*c where c is an interger. (x*x)+5 and (x*x)-5 would mean the two perfect squares would lie 10 intergers apart, and there arent any 2 consecutive perfect squares that lie an even number apart. a perfect square n^2 predicts the next perfect square more simply as n^2 + n + n + 1. you can visualize a square with n units per side. if you increase the size of the square by one, you can see how you are adding a strip of n+1 and n to the original n*n square size. based on this you would have to find non consecutive perfect squares. 1,4,9,16 shows us that 1 and 9 are 8 apart, and 4 and 16 are 12 apart. there is no pair inbetween that satisfies the original requirement.

if i am misunderstanding your question...

one possible interpretation of your wording is that a number squared and either increased or decreased means the number need not be both increased and decreased, but only one or the other. that would mean x^2+5=c^2 with c being the interger.

lets plug in anything for c equals 3. you get x^2+5=9 x^2=4, x=2. this satisfies the main equation, but i dont see how -1 could be a perfect square. you can choose these because you know that 4 - 9 = 5. 2^2 - 3^2 = 5.

if you interpret "when squared, and either increased or decreased by 5" to mean that the number is increased or decreased by 5 before the squaring... then you have (x+5) or (x-5) when either are taken to the 1/2 power (that means square root) equals a perfect square.

am i misunderstanding something? is the problem solvable?

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Check: X2 = 29/4

And X2 - 5 = 29/4 - 5 = 9/4 = (3/2)2 ... a square.

And X2 + 5 = 29/4 + 5 = 49/4 = (7/2)2 ...

I checked today and this is the right solution now can anyone find the equation so that any number will work in place of 5,-5 like 13,-13. It is a one varable equation.

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depends on what you meant by perfect square( i thought it was an integer squared)

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AKG

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What kind of limitations do we have on this

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Gokul43201

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If your difference between squares is some N, solutions are

[tex] a = \frac {x + N/x} {2} [/tex]

[tex] b = \frac {x - N/x} {2} [/tex]

for all x (in the domain of your choice, since this is not clear yet)

Integral pair solutions exist for all differences that are odd or multiples of 4 (and for no others).

Edit : Misunderstood question. Actually, need solutions for the requirement that adding "as well as" subtracting N give rational squares.

Will rethink.

[tex] a = \frac {x + N/x} {2} [/tex]

[tex] b = \frac {x - N/x} {2} [/tex]

for all x (in the domain of your choice, since this is not clear yet)

Integral pair solutions exist for all differences that are odd or multiples of 4 (and for no others).

Edit : Misunderstood question. Actually, need solutions for the requirement that adding "as well as" subtracting N give rational squares.

Will rethink.

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Gokul43201

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Hence, 49/4 and 9/4 are the outer squares and the solution is the square root of the mean, ie : sqrt(29)/2

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i always thought perfect squares had to be squares of integers?

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They are, seems there is a misunderstanding here.

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Gokul43201

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Gokul43201 said:

Hence, 49/4 and 9/4 are the outer squares and the solution is the square root of the mean, ie : sqrt(29)/2

Another solution is found by looking for a difference of 160 between integer squares. This is found for 9 and 169. Hence 9/16 and 169/16 are the outer squares. So sqrt(89)/4 is another solution.

An infinite number of such solutions can be found, because there are an infinite number of multiples of GCD(10,8), such that those multiples are 10 times a perfect square.

eg : you can always find squares that have differences of 40, 160, 360, 640, ...10*(2n)^2 and they are ({9,49},{36,196},{81,441},{144,784}...{9n^2,49n^2})

So that's your general solution...at least for 10. (maybe)You can generalize it further for any N.

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Gokul43201

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I am anxious to figgure this out.

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Gokul43201

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scox said:

I am anxious to figgure this out.

I gave you the answer(s) in my previous posts.

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