# A 1964 Puzzle

1. Oct 29, 2006

### K Sengupta

Determine all possible integer solutions (p,q) of the equation:

sqrt(p+ sqrt(p+ sqrt(...(p + sqrt(p))...))) = q

The "sqrt" symbol in the above relationship is repeated 1964 times.

Note: sqrt(x) stands for the square root of x.

2. Oct 29, 2006

### Office_Shredder

Staff Emeritus
Try squaring both sides

3. Oct 29, 2006

### 0rthodontist

I don't see an immediate way to do it, but by testing in Haskell I can tell you that the answer will probably be the set of all (x^2-x, x) for non-negative integer x except for the pair (0, 1) which is not present. Unless my results are misleading due to rounding error.

Edit: actually now I'm thinking the only solution is (0, 0). If x is not an integer and p is an integer, then sqrt(p + x) is not an integer. Therefore by induction if any of the terms within any of the square root signs are not integers, q is not an integer. So if q is an integer, then
sqrt(p + sqrt(p)) is integer. But that requires that p is a perfect square, so p = k^2 for some k. Then p + sqrt(p) = k + k^2 = k(k+1) is also a perfect square. The only non-negative integer that satisfies that is k = 0.

Last edited: Oct 29, 2006
4. Oct 29, 2006

### K Sengupta

On Today 03:13 PM; Office_Shredder wrote:

PHP:
Try squaring both sides
I am a little confused.

If the relationship :
sqrt(p+ sqrt(p+ sqrt(...(p + sqrt(p))...))) = q ........(*)
was inclusive of the repetition of the "sqrt" symbol an infinite number of times; one would readily have obtained p = q^2 - q; giving an infinite number of solutions to the problem for an integer p.

However, it may be noted that in terms of the problem, the "sqrt" symbol is repeated precisely 1964 times in (*), and consequently I am not very sure that the parametric relationship p = q^2 - q; would satisfy the conditions of the problem.

5. Oct 29, 2006

### arildno

$$2=\sqrt{4}=\sqrt{2+2}=\sqrt{2+\sqrt{2+2}}=\sqrt{2+\sqrt{2+\sqrt{2+2}}}$$

6. Oct 29, 2006

### K Sengupta

On Today at 03:28 PM; 0rthodontist wrote:

PHP:
........the only solution is (0, 0).
Your result completely agrees with mine. Incidentally, I pursued the same proceedure to solve the problem.

It does seem that for k(k+1) to be a perfect square in any given iteration; k must correspond to 0 in the same iteration.

Last edited: Oct 29, 2006
7. Oct 29, 2006

### StatusX

8. Oct 29, 2006

### K Sengupta

On Today 08:49 PM; StatusX wrote:

PHP:
True. I confirm having obviated the final paragraph of my previous post in
conformity with the foregoing.

9. Oct 29, 2006

### K Sengupta

On Today 08:12 PM; arildno wrote:

PHP:
2 = sqrt(4) = sqrt (2+2) = sqrt(2+ sqrt(2+2)) = sqrt(2+ sqrt(2+sqrt(2+2))) = .......
Your solution corresponds to the equation:
sqrt(p+ sqrt(p+ sqrt(...(p + sqrt(p+p))))) = q; where “sqrt” symbol is repeated 1964 times……………………..(#)

rather than the original problem.

We consider the sequence S_1(p) = S_1(p) = sqrt(p + p); and
S_(i+1)(p) = 2+ sqrt(S_i(p)) for i = 1,2,…..,1963.

For p=0 and 2, we readily observe that S_(i+1)(p) = S_i(p); for all non-negative integers i.
So, S_(1964)(p) = q; yields (p,q) = (0,0) and (2,2) as two of the integer solutions to (#).

I thank you for the new problem and, if possible I will try to obtain other values for (p,q) or try to prove the non-existance of any other integer solutions to (#).

Last edited: Oct 29, 2006
10. Oct 29, 2006

### Office_Shredder

Staff Emeritus
You're right, the ... in the sqrt confused me. Sorry about that

11. Oct 29, 2006

### 0rthodontist

Except for the case 0 = 1^2 - 1--(0, 1) does not satisfy that infinite equation.

12. Oct 30, 2006

### K Sengupta

On Today at 12:04 AM; 0rthodontist wrote:

PHP:
Except for the case 0 = 1^2 - 1--(0, 1) does not satisfy that infinite equation.
True. I stand corrected.

The parametric solution p=q^2-q, for the infinite equation:
q = sqrt(p+(sqrt(p+…..sqrt(p+sqrt(p))))); where both p and q are integers and; the “sqrt” symbol is repeated 1964 times………….(i)
suggests that p must be non-negative.

Otherwise, for any given negative p, sqrt(p) is never a real number, which is a contradiction.

Since from (i) , q>= sqrt(p); for non-negative p; it follows that q must be non-negative.
Accordingly, we only need to consider only the non-negative roots of the equation:
q^2 – q = p.

Now, q^2-q –p =0, gives:
q = (1+/- sqrt(4p+1))/2
For p>=1; q = (1- sqrt(4p+1))/2, is negative which is a contradiction.
So, q = (1+sqrt(4p+1))/2; for all p>=1 …….(ii)

For p=0, we obtain q^2 = q; so that, q = 0,1.

A direct check with equation (i) would reveal that:
p=0 yields q=0 and (p,q) = (0,1) is a contradiction.

In relationship (ii); the minimum value of p >=1 for which sqrt(4p+1) is an integer occurs at p=2; giving:
q = (1+3)/2=2 as the minimum value of q whenever p>=1.

Subsequently; the required integer solution for the infinite equation (i) would be:
(p,q) = (0, 0) and:
p= q^2-q; whenever q>=2.

13. Oct 30, 2006

### K Sengupta

On Today at 08:12 PM; arildno wrote:

PHP:
2 = sqrt(4) = sqrt (2+2) = sqrt(2+ sqrt(2+2)) = sqrt(2+ sqrt(2+sqrt(2+2))) = .......
arildno‘s problem requires one to determine all possible integer solutions to the undernoted equation:

sqrt( p+sqrt(p+(……+sqrt(p+sqrt(2p))…)).= q; where both p and q are integers and the “sqrt” symbol is repeated 1964 times…………..(#)

We observe that:
S_1(p) = sqrt(2p);
S_2(p) = sqrt(p+sqrt(2p));
S_3(p) = sqrt( p+ sqrt(p+sqrt(2p))
S_(i+1)(p) = sqrt( p+ sqrt(S_i(p))); for i = 2,3,…,1963.

Now, S_1(p) is an integer only when p =2* k^2, for some integer k.
Accordingly, S_2(p) = sqrt(2k(k+1)).
Clearly, for q to correspond to an integer, it follows that S_2(p) must be an integer; so that:
2k(k+1) = (S_2(p))^2

Solving the underlying Pell’s equation for k<300; we obtain:
(k,p,S_1(p), S_2(p), S_3(p))
= (1,2,2,2,2); (8,128, 16, 12, sqrt(140)); (49, 4802, 98, 70, sqrt(4872));
(288, 165888, 576, 408, sqrt(166296))

Now,
S_1(p) = S_i(p) = 0, whenever p=0; for i=2,3,….,1964
S_1(p) = S_i(p) = 2, whenever p=2; for i = 2,3,…,1964

Since S_3(p) does not correspond to an integer for p< 2*300^2 = 180,000; we can safely conclude that the equation (#) admits of the solution:
(p,q) = (0,0); (2,2) whenever p<180,000.

14. Oct 30, 2006

### K Sengupta

On Today at 08:12 PM; arildno wrote:

PHP:
2 = sqrt(4) = sqrt (2+2) = sqrt(2+ sqrt(2+2)) = sqrt(2+ sqrt(2+sqrt(2+2))) = .......
The above expression suggests another problem which correspond to the generalized version of the previous problem.

This problem would require one to find all possible integer triplets (p,q,r) satisfying:

sqrt( p+sqrt(p+(……+sqrt(p+sqrt(p + q)))= r; where the “sqrt” symbol is repeated 1964 times
....…………..(A)

Till date, I have not been able to derive a satisfactory solution to (A) other than:
(p,q,r) = (2,2,2) and (0,0,0).

Consequently, I am looking for an analytic method which may lead to a solution to the foregoing problem.

Last edited: Oct 30, 2006