Do people actually attempt to solve this kind of puzzle manually?

[musicgold]

Homework Statement

This is not a homework problem. I was trying to solve a puzzle from a book : 536 curious problems and puzzles. Puzzle no. 199 in the attached picture.

Relevant Equations

The three terms should look like:
$2 a_1 + d , ~~ 2 a_1 + 2d , ~~ 2a_1 + 3d$

After trying a few combinations of a1 and d, I looked up the answer and the anwer was: 482, 3362, 6242, where d is 2880.

After seeing such a complicated answer a few questions popped into my head:

1. Could someone have actually manually worked this puzzle out or did the author just ran a simulation to find such unique facts to create a puzzle?
2. Did people in the time before logarithms actually attempt to solve such problems? If yes, how would they go about it?

P.S. I have read a book on Fermat's Last Theorem and know that people did work for years and years on such problems, but would love to get some additional perspective. It would be great if you could refer me to a book or site that discusses such ideas.

Thanks

 

[Vanadium 50]

In general, if you are thinking "three elements of an arithmetic series" start with a-b, a, a +b. Now your squares are 2a - b, 2a and 2a + b. a is even. I'd start from there.

The book was published after logarithms were invented. How do i know? It's not written in Shakespearean English.

 

[Vanadium 50]

Let me add that I have done this problem now. Yes, my starting point will get you there, but the path is not short or pretty or terribly enlightening. Lots of algebra - you can't be afraid to dive right in and you can't be afraid to change variables. Twice.

Forsooth!

 

[Vanadium 50]

As a final PS, there are multiple solutions, e.g. 4562, 20402, and 36242, with square roots of pairwise sums of 158, 202 and 238.

My favorite solution is 2,2,2.

 

[DaveE]

In general, if you are thinking "three elements of an arithmetic series" start with a-b, a, a +b. Now your squares are 2a - b, 2a and 2a + b. a is even. I'd start from there.

The book was published after logarithms were invented. How do i know? It's not written in Shakespearean English.

How did you know a is even?

 

[Vanadium 50]

If 2a is a perfect square, a must contain the other factor of 2 and thus be even.

 

[Vanadium 50]

And, just for fun,

Prove or find a counterexample that every number that forms a solution ends in 2.

 

[musicgold]

Let me add that I have done this problem now. Yes, my starting point will get you there, but the path is not short or pretty or terribly enlightening. Lots of algebra - you can't be afraid to dive right in and you can't be afraid to change variables. Twice.

Forsooth!

Lots of algebra - that was my main problem in the first place; I couldn't think of any algebraic technique to tackle this problem. All I could do was to try various combinations a and b, that too using a calculator.

Could you give me more hints?
Thanks

 

[Vanadium 50]

I couldn't think of any algebraic technique to tackle this problem.

Substitute. p²+q² and 2pq might have some utility.

 

[PeroK]

Lots of algebra - that was my main problem in the first place; I couldn't think of any algebraic technique to tackle this problem. All I could do was to try various combinations a and b, that too using a calculator.

Could you give me more hints?
Thanks

Here are some ideas. What we are looking for is three evenly spaced numbers that are all squares. It's not hard to see that $1, 25, 49$ satisfies this. But, the problem is set up so that the middle number is $2a$ so must be even. So that one doesn't workl. A quick check shows that there doesn't appear to be a small solution with the middle number even. So, let's use some algebra.

Let the middle number be $n^2$, the large number be $(n + m)^2$ and the lower number be $[n - (m + r)]^2$. It's an exercise for you to explain why we have $m + r$ here. How do I know this number is larger than $m$? $r$ is the key here. Note also that $n$ must be even.

Now we subtract these squares to give the common difference, which simplifies to:

$2nr - r^2 = 2m^2 + 2mr$

This equation cannot be satified if $r$ is odd. So, let's start trying some even $r$. Over to you.

 

[PeroK]

PS The first solution I get out is $4, 100, 196$, which corresponds to $a = -92, d = 96$. But, the problem requires $a \ge 0$, so that criterion needs to be included somewhere.

 

[pasmith]

If the three numbers are A - d, A, A + d then the pairwise sums are 2A - d, 2A, 2A + d. We can set 2A = k^2 and <br /> \begin{align*}<br /> 2A - d &amp;= (p - q)^2, \\<br /> 2A + d &amp;= (p + q)^2. <br /> \end{align*}<br /> Now because the difference between 2A + d and 2A - d is even, we have that (p +q)^2 and (p - q)^2 and hence p + q and p - q themselves are either both even or both odd. It follows that their sum (2p) and difference (2q) are both even, so p and q are integers with p &gt; q.

Subtracting these yields d = 2pq. Both of these equations then reduce to <br /> k^2 = p^2 + q^2. If A - d &gt; 0 then <br /> p^2 + q^2 &gt; 4pq which will be satisfied if p &gt; 4q. So the problem is to find a pythagorean triple (q, p ,k) with p &gt; 4q and k even. (If we find one with k odd, then we can use (2q, 2p, 2k).)

 

[musicgold]

I am still stuck. Here is my attempt:

Let the three numbers be $A−d, ~ A, ~A+d$ then their pairwise sums are
$2A−d, ~ 2A, ~ 2A+d$.

Let's assume that the three square numbers are $p^2, ~ k^2, ~ q^2$, respectively.

Then I used $A = \frac {k^2} {2}$ to get the following values of $d$.
$d = k^2 - p^2$
$d = q^2 - k^2$

Equating the two values of d gave me
$q^2 +p^2 = 2 k^2$

Adding $pq$ terms on both sides to get perfect squares, we get
$(q+p)^2 = 2k^2 + 2 pq$
$(q-p)^2 = 2k^2 - 2pq$

I don't know how to proceed from here to be able to guess A and d.

I see that my solution is very similar to to that of pasmith, but I don't follow how he is assuming $d = 2pq$.

Thanks

 

[Vanadium 50]

You're struggling because your plan to solve this is not a good one. You have two equations in three unknowns, and the solutions are integers. Your plan is to try and solve two equations in three unknowns, ignoring the fact that the numbers are integers, and hoping that when the numbers pop out - which they won't (2 equations, 3 unknowns - they will be integers.

This is not a good plan.

 

[musicgold]

pasmith, could you explain me how you can make the following assumption?

<br /> \begin{align*}<br /> 2A - d &amp;= (p - q)^2, \\<br /> 2A + d &amp;= (p + q)^2.<br /> \end{align*}<br />

 

[Vanadium 50]

You can always write two numbers A and B as p + q and p - q.

But you're back to your same non-working plan.

2,2,2
482, 3362, 6242
4562, 20402, 36242
39042, 272322, 505602

are all solutions. Indeed, there are an infinite number of solutions. So you're plan of having two equations in three unknowns miraculously popping out a solution is not a good one. Nor is hoping for a second miracle that the number is an integer.

The path you are on will not take you to a solution no matter how hard you work at it. You need another one.

 

[PeroK]

pasmith, could you explain me how you can make the following assumption?

I think this problem is quite hard. You might like first to try the easier problem I posted in post #10 just to find an equation for all squares in an arithmetic progression. That was my approach, although I didn't find a way to filter my solutions to the ones that fit this problem. But, in a way, it's a nicer problem to find them all.

 

[pasmith]

pasmith, could you explain me how you can make the following assumption?

2A - d and 2A + d are required to be squares. So set 2A - d = a^2 and 2A + d = b^2. Subtracting one from the other gives 2d = b^2 - a^2 = (b + a)(b - a). Now at least one of b - a and b + a is even (because the product is even), which means that the other is also (because b + a = (b - a) + 2a). Hence we may set 2q = b - a, \qquad 2p = b + a so that a = p - q, \qquad b = p + q.

 

[musicgold]

The path you are on will not take you to a solution no matter how hard you work at it. You need another one.

Can you please point me to a webpage or book where I can read more on the approach you are using?

 

[musicgold]

Let the middle number be $n^2$, the large number be $(n + m)^2$ and the lower number be $[n - (m + r)]^2$. It's an exercise for you to explain why we have $m + r$ here. How do I know this number is larger than $m$? $r$ is the key here. Note also that $n$ must be even.

Now we subtract these squares to give the common difference, which simplifies to:

$2nr - r^2 = 2m^2 + 2mr$

This equation cannot be satified if $r$ is odd. So, let's start trying some even $r$. Over to you.

Two questions:

1. If the pairwise sums are equidistant from each other on the number line, why are we assuming a strange term like $(m + r)$ in defining them? Is this kind of shortcut to simplify calculations?

2. I am able to reach up to this step: $2nr - r^2 = 2m^2 + 2mr$. However, again, I don't know how to proceed from here. Now I have 3 variables to guess: n, m and r. In my initial brute force approach, I was dealing with only two variables - the initial term and the difference d. What am I missing?

Thanks for bearing with me.

 

[PeroK]

Two questions:

1. If the pairwise sums are equidistant from each other on the number line, why are we assuming a strange term like $(m + r)$ in defining them? Is this kind of shortcut to simplify calculations?

2. I am able to reach up to this step: $2nr - r^2 = 2m^2 + 2mr$. However, again, I don't know how to proceed from here. Now I have 3 variables to guess: n, m and r. In my initial brute force approach, I was dealing with only two variables - the initial term and the difference d. What am I missing?

Thanks for bearing with me.

1. Take an example like $1, 25, 49$ which is $1^2, 5^2, 7^2$. The difference between the larger pair $7 - 5$ must be greater than the difference in the smaller pair $5 - 1$. In this case $n = 5, m = 2$ and $r=2$.

You can show this in general. Suppose we have $a^2, b^2, c^2$, then $c^2 - b^2 = (c-b)(c+b) = b^2 - a^2 = (b-a)(b+a)$. We know that $c + b > b + a$, so we must have $c - b < b - a$.

2. Now you have all the solutions. It's not a question of brute force. You can choose almost any $r, m$ and you get a solution for $n$. Note that $r$ must be even. There are an infinite number of them. For example:

$r = 2, m = 1$ gives $1, 25, 49$

$r = 2, m = 2$ gives $7, 13, 17$

And you can generate as many as you want.

Note: unfortunately, I can't see a way to restrict this full set to just the ones we want for the given problem.

 

[Vanadium 50]

There is only so much floundering one can stand.

With sufficient substitution, one can determine the sequence is (p^2+q^2)^2/2 +/- or 0 4pq(p^2-q^2) with p and q being positive integers, either both even or both odd. Without loss of generality, p > q.

We want all three terms to be non-negative, so (p^2+q^2)^2 &gt; 8pq(p^2-q^2)

OK, so what's the smallest q can be? One.
Can p be 3? No, (p^2+q^2)^2 &lt; 8pq(p^2-q^2)
5? Same problem. 7? Same problem. 9 works. That gives us 482-3362-6242
From here, we can keep plugging in: each value of p gives a new sequence: 2162-7442-1272 for 11, then 5714-14450-23186 for 13, etc.

Next smallest? Two.
Then the smallest q can be is 16, and that gets us 1544-33800-66056. Eighteen gets us 7712-53792-99872, and so on forever.

Next smallest? Three.
The first q that works is 23, for 1202-144722-288242.

This is a way to systematically generate as many solutions as one wants.

 

[musicgold]

There is only so much floundering one can stand.

With sufficient substitution, one can determine the sequence is (p^2+q^2)^2/2 +/- or 0 4pq(p^2-q^2) with p and q being positive integers, either both even or both odd. Without loss of generality, p > q.

We want all three terms to be non-negative, so (p^2+q^2)^2 &gt; 8pq(p^2-q^2)

OK, so what's the smallest q can be? One.
Can p be 3? No, (p^2+q^2)^2 &lt; 8pq(p^2-q^2)
5? Same problem. 7? Same problem. 9 works. That gives us 482-3362-6242
From here, we can keep plugging in: each value of p gives a new sequence: 2162-7442-1272 for 11, then 5714-14450-23186 for 13, etc.

Next smallest? Two.
Then the smallest q can be is 16, and that gets us 1544-33800-66056. Eighteen gets us 7712-53792-99872, and so on forever.

Next smallest? Three.
The first q that works is 23, for 1202-144722-288242.

This is a way to systematically generate as many solutions as one wants.

Thank you. Happy Teacher's Day to all my teachers.