# Integer Solutions of sqrt(p) = q (1964 times)

• K Sengupta
In summary, the given equation has two possible integer solutions: (0, 0) and (2, 2). This is determined by solving the underlying Pell's equation and considering the properties of the sequence S_i(p) defined in the given equation. For any other non-negative integer values of p, there is no corresponding integer value for q.
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.

Try squaring both sides

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.

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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.

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

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.

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K Sengupta said:
However, I was also unable to conclusively prove that no other solution in addition to (0,0) satisfy conditions of the given problem.

Orthodontist already did.

On Today 08:49 PM; StatusX wrote:

PHP:
Orthodontist already did.

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

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 (#).

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K Sengupta said:
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.

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

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

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.

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.

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.

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## 1. What are integer solutions?

Integer solutions refer to numbers that are whole numbers, either positive or negative, and do not contain any fractions or decimals.

## 2. What is the meaning of sqrt(p) = q?

The notation sqrt(p) = q means that the square root of p (represented by the symbol √) is equal to q. In other words, q is the number that, when multiplied by itself, equals p.

## 3. How many integer solutions are there for sqrt(p) = q (1964 times)?

There are an infinite number of integer solutions for this equation. This is because when you multiply a number by itself 1964 times, you will always get a perfect square, which can have an infinite number of square roots.

## 4. Can you give an example of an integer solution for sqrt(p) = q (1964 times)?

One possible integer solution for this equation is sqrt(1964) = 44. This is because 44 multiplied by itself 1964 times equals 1964, which is a perfect square.

## 5. How can you find the integer solutions for sqrt(p) = q (1964 times)?

To find the integer solutions for this equation, you can use a calculator to take the square root of p and then multiply it by itself 1964 times. Alternatively, you can use algebraic methods to solve the equation and find the integer solutions.

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