Interesting Numeric Sets

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Yarem4uk

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Plese, help me with this problem.
a) The finite or infinite is the set of four natural numbers such that the product of any two numbers from each such four plus 1, is a square of natural number?
b) Specify at least one set of five positive rational numbers such as that neither is a natural number and the product of any two of these numbers, plus 1 is the square of a rational number.
Sorry for my bad english.

I thought, if the set of four natural numbers a<b<c<d, so i have to show, that
ab+1
ac+1
ad+1
bc+1
bd+1
cd+1
are a square of natural number.
For any m>2 or m=2 if
a=m-1
b=m+1
c=4m
d= mk-1(mk-1) k Є N

ab+1=m2
ac+1=(2m-1)2
bc+1=(2m+1)2
cd+1=(2mk-1)2
...
But ad+1 and bd+1... I don't know. It is all, that i can to do.
Thanks for any help.
 
I'm not 100% sure I understand the problem. Is your goal to determine if there are infinitely many sets of four numbers that satisfy these criteria? Or is it just to find one specific example of a set of four numbers that works? If you need only one example, it should be trivial to find a specific set.

On the other hand, if you need to show there are infinitely many possible sets, I'd think about fixing \(d\) to be a real number. You've given \(a,\: b, \:\) and \(c\) in terms of another arbitrary integer \(m\), so even if you say, for example, \(d = 7\), you still have infinitely many solutions (provided of course that your value of \(d\) works). Can you see why that is? Then can you think of a real number \(d\) that can guarantee \(ad + 1\), \(bd + 1\) and \(cd+1\) are all the square of a natural number?
 
Thank you!
My goal is to determine if there are infinitely many sets of four integer positive numbers that satisfy these criteria.
I found a lot of such sets. For example
a=1, b=3, c=8, d=120
a=2, b=4, c=12, d=420
a=3, b=5, c=16, d=1008...
And I think, that the set of such four numbers is infinite. But i don't know, how to prove it. I tried to find some formulas for the a, b, c and d with the using integer positive m, but it doesn't work.
Sory for my english.
 
I have NOT solved this, and I am running out of time to work on it. But I have progressed.

Given the numeric exploration already done by the OP, his three examples of quadruples prove at least a finite number of such quadruples exist. And by extending the scope of the problem to non-negative integers, I found by trivial numeric exploration a fourth example, namely (0, 2, 4, 12) because

0 * 2 + 1 = 1, a perfect square
0 * 4 + 1 = 1, a perfect square
0 * 12 + 1 = 1, a perfect square
2 * 4 + 1 = 9, a perfect square
2 * 12 + 1 = 25, a perfect square
4 * 12 + 1 = 49, a perfect square.

Moreover, based on the OP's examples, it is easy to deduce a pattern for the three numbers in a triple, namely

[MATH]i \in \mathbb Z^+.[/MATH]
[MATH]\therefore i + 2 \text { and } 4i + 4 \in \mathbb Z^+.[/MATH]
[MATH]i(i + 2) + 1 = i^2 + 2i + 1 = (i + 1)^2.[/MATH]
[MATH]i(4i + 4) + 1 = 4i^2 + 4i + 1 = (2i + 1)^2.[/MATH]
[MATH](i + 2)(4i + 4) + 1 = 4i^2 + 12i + 9 = (2i + 3)^2.[/MATH]
So there are an infinite number of triples.

By my extension, I have found a pattern that provides loads of further numeric examples for quadruples, namely

[MATH]16i^3 + 48i^2 + 44i + 12 \implies f(0) = 12,\ f(1) = 120, \ f(2) = 420[/MATH]
[MATH]\text {and } f(3) = 1008.[/MATH]
But I doubt I shall have time to try to figure out whether

[MATH]i \in \mathbb Z \text { and } i \ge 1 \implies[/MATH]
[MATH]\sqrt{i * f(i) + 1} = \sqrt{16i^4 + 48i^3 + 44i^2 + 12i + 1} \in \mathbb Z,[/MATH]
[MATH]\sqrt{(i + 2) * f(i) + 1} \in \mathbb Z \text { and}[/MATH]
[MATH]\sqrt{(4i + i)) * f(i) + 1} \in \mathbb Z.[/MATH]
I shall try finding the time to continue working on this but am not at all sure that I can.
 
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Technically the problem text never specifies that the condition \(a < b < c < d\) is necessary. In fact, it doesn't even explicitly say that the four numbers need to be unique. Therefore, if 0 is considered to be a natural number, there's infinitely many trivial solutions of the form \( (m, 0, 0, 0) \) for any natural number \(m\).

Finding non-trivial solutions, however, is proving much more difficult than I initially thought.
 
ksdhart2 said:
Technically the problem text never specifies that the condition \(a < b < c < d\) is necessary. In fact, it doesn't even explicitly say that the four numbers need to be unique. Therefore, if 0 is considered to be a natural number, there's infinitely many trivial solutions of the form \( (m, 0, 0, 0) \) for any natural number \(m\).

Finding non-trivial solutions, however, is proving much more difficult than I initially thought.
Click to expand...
I have been assuming that all four numbers must be different. If that is true, you can assume that a < b < c < d without loss of generality.

I have reason to believe from some very limited numeric experimentation that

[MATH]\{i,\ i + 2,\ 4i + 4, \text { and } 16i^3 + 48i^2 + 44i + 12\} \text { if } i \in \mathbb Z^+[/MATH]
describes the set desired, and obviously that set is infinite.

So the only step remaining is to factor three quartics, but I hate factoring and have no time for the next three days.

EDIT: There is a plausibility argument that a cubic is a reasonable function. When multiplied by the other candidates i, i + 2, and 4i + 4, you get a quartic, which could be the square of a perfect square. It was to find the coefficients of the cubic candidate that I needed to find the (0, 2, 4, 12) example. Of course this is all hand waving, but it may not be silly hand waving.
 
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ksdhart2 said:
Technically the problem text never specifies that the condition \(a < b < c < d\) is necessary. In fact, it doesn't even explicitly say that the four numbers need to be unique. Therefore, if 0 is considered to be a natural number, there's infinitely many trivial solutions of the form \( (m, 0, 0, 0) \) for any natural number \(m\).

Finding non-trivial solutions, however, is proving much more difficult than I initially thought.
Click to expand...
Thank you!
I am from Ukraine, and in my country 0 is not a natural number.
And, for example, if some two numbers are unique (a=b), then a*b+1 = a2+1 have to be a square of natural number, but it's not possible.
 
JeffM said:
I have NOT solved this, and I am running out of time to work on it. But I have progressed.

Given the numeric exploration already done by the OP, his three examples of quadruples prove at least a finite number of such quadruples exist. And by extending the scope of the problem to non-negative integers, I found by trivial numeric exploration a fourth example, namely (0, 2, 4, 12) because

0 * 2 + 1 = 1, a perfect square
0 * 4 + 1 = 1, a perfect square
0 * 12 + 1 = 1, a perfect square
2 * 4 + 1 = 9, a perfect square
2 * 12 + 1 = 25, a perfect square
4 * 12 + 1 = 49, a perfect square.

Moreover, based on the OP's examples, it is easy to deduce a pattern for the three numbers in a triple, namely

[MATH]i \in \mathbb Z^+.[/MATH]
[MATH]\therefore i + 2 \text { and } 4i + 4 \in \mathbb Z^+.[/MATH]
[MATH]i(i + 2) + 1 = i^2 + 2i + 1 = (i + 1)^2.[/MATH]
[MATH]i(4i + 4) + 1 = 4i^2 + 4i + 1 = (2i + 1)^2.[/MATH]
[MATH](i + 2)(4i + 4) + 1 = 4i^2 + 12i + 9 = (2i + 3)^2.[/MATH]
So there are an infinite number of triples.

By my extension, I have found a pattern that provides loads of further numeric examples for quadruples, namely

[MATH]16i^3 + 48i^2 + 44i + 12 \implies f(0) = 12,\ f(1) = 120, \ f(2) = 420[/MATH]
[MATH]\text {and } f(3) = 1008.[/MATH]
But I doubt I shall have time to try to figure out whether

[MATH]i \in \mathbb Z \text { and } i \ge 1 \implies[/MATH]
[MATH]\sqrt{i * f(i) + 1} = \sqrt{16i^4 + 48i^3 + 44i^2 + 12i + 1} \in \mathbb Z,[/MATH]
[MATH]\sqrt{(i + 2) * f(i) + 1} \in \mathbb Z \text { and}[/MATH]
[MATH]\sqrt{(4i + i)) * f(i) + 1} \in \mathbb Z.[/MATH]
I shall try finding the time to continue working on this but am not at all sure that I can.
Click to expand...
Thank you very much for your work!
It is a big problem for me to find a correct formula for the fourth number. Today I tried to use a computer to have more quartic.
Here's what I got.
 

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Actually, the factoring's really not that bad, despite them being quartics. I'll do the first one and leave the remainder to the OP to factor. Let's first write out each of the three things we need to factor:

\(\displaystyle
ad + 1 = (m)(16m^3 + 48m^2 + 44m + 12) + 1 = 16m^4 + 48m^3 + 44m^2 + 12m + 1 \\
bd + 1 = (m+2)(16m^3 + 48m^2 + 44m + 12) + 1 = 16m^4 + 80m^3 + 140m^2 + 100m + 25 \\
cd + 1 = (4m+4)(16m^3 + 48m^2 + 44m + 12) + 1 = 64m^4 + 256m^3 + 368m^2 + 224m + 49
\)
Next we'll make use of a substitution-based method to "depress" these quartics. Given a generic quartic \(\alpha m^4 + \beta m^3 + \gamma m^2 + \delta m + \epsilon\), we use the substitution \(\displaystyle m = y - \frac{\beta}{4 \alpha}\). For the first specific quartic here, we'll use \(m = y - \frac{48}{4(16)} = y - \frac{3}{4} \). Then:

\(\displaystyle 16m^4 + 48m^3 + 44m^2 + 12m + 1 = 16 \left( y - \frac{3}{4} \right)^4 + 48\left( y - \frac{3}{4} \right)^3 + 44\left( y - \frac{3}{4} \right)^2 + 12\left( y - \frac{3}{4} \right) + 1 = 16y^4 - 10y^2 + \frac{25}{16}\)

If we make another substitution, we can see that this is really a quadratic. Let \(z = y^2\):

\(\displaystyle 16m^4 + 48m^3 + 44m^2 + 12m + 1 = 16z^2 - 10z + \frac{25}{16} = 16\left( z - \frac{5}{16} \right)^2\)

And finally, back substitute \(\displaystyle z = y^2 = \left( m + \frac{3}{4} \right)^2\)

\(\displaystyle 16\left( z - \frac{5}{16} \right)^2 = 16 \left[ \left( m + \frac{3}{4} \right)^2 - \frac{5}{16} \right]^2 = 16 \left( m^2 + \frac{3m}{2} + \frac{1}{4} \right)^2 = (4m^2 + 6m + 1)^2\)
 
I definitely shall let the OP do the other two as an exercise because all that is more algebra than I like to do.

Between the three of us, we have, I think, proved that the set is indeed infinite.
 
ksdhart2 said:
Actually, the factoring's really not that bad, despite them being quartics. I'll do the first one and leave the remainder to the OP to factor. Let's first write out each of the three things we need to factor:

\(\displaystyle
ad + 1 = (m)(16m^3 + 48m^2 + 44m + 12) + 1 = 16m^4 + 48m^3 + 44m^2 + 12m + 1 \\
bd + 1 = (m+2)(16m^3 + 48m^2 + 44m + 12) + 1 = 16m^4 + 80m^3 + 140m^2 + 100m + 25 \\
cd + 1 = (4m+4)(16m^3 + 48m^2 + 44m + 12) + 1 = 64m^4 + 256m^3 + 368m^2 + 224m + 49
\)
Next we'll make use of a substitution-based method to "depress" these quartics. Given a generic quartic \(\alpha m^4 + \beta m^3 + \gamma m^2 + \delta m + \epsilon\), we use the substitution \(\displaystyle m = y - \frac{\beta}{4 \alpha}\). For the first specific quartic here, we'll use \(m = y - \frac{48}{4(16)} = y - \frac{3}{4} \). Then:

\(\displaystyle 16m^4 + 48m^3 + 44m^2 + 12m + 1 = 16 \left( y - \frac{3}{4} \right)^4 + 48\left( y - \frac{3}{4} \right)^3 + 44\left( y - \frac{3}{4} \right)^2 + 12\left( y - \frac{3}{4} \right) + 1 = 16y^4 - 10y^2 + \frac{25}{16}\)

If we make another substitution, we can see that this is really a quadratic. Let \(z = y^2\):

\(\displaystyle 16m^4 + 48m^3 + 44m^2 + 12m + 1 = 16z^2 - 10z + \frac{25}{16} = 16\left( z - \frac{5}{16} \right)^2\)

And finally, back substitute \(\displaystyle z = y^2 = \left( m + \frac{3}{4} \right)^2\)

\(\displaystyle 16\left( z - \frac{5}{16} \right)^2 = 16 \left[ \left( m + \frac{3}{4} \right)^2 - \frac{5}{16} \right]^2 = 16 \left( m^2 + \frac{3m}{2} + \frac{1}{4} \right)^2 = (4m^2 + 6m + 1)^2\)
Click to expand...
It is so cool! Thank you!
bd+1=(4m2+10m+5)2
cd+1=(8m2+16m+7)2
I worked with this problem so much time and now I feel great!
 
There is a second part in this task.

Find such five different rational numbers (any from this numbers is not a natural number), that the product of any two numbers from each such five plus 1 is a square of rational number.
Sorry for my arrogance. I haven't written this part before, because I have no ideas here...
 
I can give you ideas but no work as of now.

First, what you did in the first part for Z applies to Q so you are just looking for the formula for the fifth number in each n-tuple.

Second, I do not know whether it is possible to solve this expanded problem in Z, but if you can, that solution will apply to Q.

Third, I might start with numerical experimentation as you did before. Use my gimmick of starting with zero rather than 1 because it will give you an easier starting point. You might want to work with multiples of 1/2 and 1/5 because they have such friendly decimal expansions.

Fourth, if the formula for the fifth number is also a cubic (which is a pure guess), you will need FIVE examples to compute the cubic's coefficients.

I may check in later tonight.

There probably is a neat way to solve this using induction, but you will need someone who actually knows math, which is not me, to go that route. I go by guess and by God.
 
JeffM said:
Fourth, if the formula for the fifth number is also a cubic (which is a pure guess), you will need FIVE examples to compute the cubic's coefficients.
Click to expand...
The above is incorrect. You will need four examples if a cubic works.
 
JeffM said:
I can give you ideas but no work as of now.

First, what you did in the first part for Z applies to Q so you are just looking for the formula for the fifth number in each n-tuple.

Second, I do not know whether it is possible to solve this expanded problem in Z, but if you can, that solution will apply to Q.

Third, I might start with numerical experimentation as you did before. Use my gimmick of starting with zero rather than 1 because it will give you an easier starting point. You might want to work with multiples of 1/2 and 1/5 because they have such friendly decimal expansions.

Fourth, if the formula for the fifth number is also a cubic (which is a pure guess), you will need FIVE examples to compute the cubic's coefficients.

I may check in later tonight.

There probably is a neat way to solve this using induction, but you will need someone who actually knows math, which is not me, to go that route. I go by guess and by God.
Click to expand...
I made numerical experimentation on my computer for the four numbers, that I know. But i could not find any fifth number (my computer can do it only for fifth number less than 109). I think, that it is impossible.
 
Yarem4uk said:
I made numerical experimentation on my computer for the four numbers, that I know. But i could not find any fifth number (my computer can do it only for fifth number less than 109). I think, that it is impossible.
Click to expand...
I am at the seashore with limited computer power, but my very cusory experimentaion showed no hits even with a = 0.

I have no clue how to prove impossibilty.
 
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