Find a 4-Digit Square Phone Number | Add 1 to Each Digit for Another Square

[xholicwriter]

find a 4-digit phone number, given that it is a square. If we add 1 unit to each digit of the phone number, the new number we obtained is also a square.

 

[Char. Limit]

So we are dealing with a number, n^2, and we are given that n<100. Our job, I'm guessing is to find an integer m such that n^2 + 1111 = m^2. Now, I could just test all n<100, but that might be too simple... although I don't know any other ways to do it!

Considering that this is a Diophantine equation, it might be better placed in the Number Theory forum, but I like it here.

BTW, computation (via Wolfram-Alpha) gives an immediate answer, but I won't reveal it here, as Wolfram-Alpha is no fun.

 

[xholicwriter]

you're on the right path, but don't use guess and check. there is a logic way to do it

 

[Char. Limit]

Oh, I see now. Starting with the equation n² + 1111 = m², we can subtract n² from both sides and then factor to obtain 1111=(m+n)(m-n). Now, since m and n are integers, m+n and m-n must also be integers, and therefore factors of 1111. 1111 has only two factors: 11 and 101. Therefore, m+n and m-n must equal 101 and 11, respectively. (assuming that both m and n are positive, and that m>n, which is obvious from the original problem)

So now we have two equations: m+n=101 and m-n=11. Two distinct equations and two unknowns mean that there is exactly one solution. Let's add the two equations to get 2m=112. Therefore, m=56. Substitute m=56 into m-n=11 to get 56=n+11 and therefore n=45.

Finally, square n to get the original number, n², which is 2025 and the 4-digit number that solves the problem.

 

[xholicwriter]

Cool! if you like this type of problem, try the next one/ find a 5-digit number. If we take off the last 3 digits, the new number we obtained is the cube root of the original number

 

[micromass]

Oh, I see now. Starting with the equation n² + 1111 = m², we can subtract n² from both sides and then factor to obtain 1111=(m+n)(m-n). Now, since m and n are integers, m+n and m-n must also be integers, and therefore factors of 1111. 1111 has only two factors: 11 and 101. Therefore, m+n and m-n must equal 101 and 11, respectively. (assuming that both m and n are positive, and that m>n, which is obvious from the original problem)

So now we have two equations: m+n=101 and m-n=11. Two distinct equations and two unknowns mean that there is exactly one solution. Let's add the two equations to get 2m=112. Therefore, m=56. Substitute m=56 into m-n=11 to get 56=n+11 and therefore n=45.

Finally, square n to get the original number, n², which is 2025 and the 4-digit number that solves the problem.

Very nice char! This also shows in general that the curve

$$x^2-y^2=1111$$

has only finitely many integer solutions. Can we also find the rational solutions? Yes! Let's take an arbitrary line through (56,45) with rational slope t. Such a line has equation

$$y=t(x-56)+45$$

The intersection of this line with the curve $x^2-y^2=1111$ will yield a rational point on the curve (furthermore, I don't think it's difficult to show that this yields all the rational points on the curve). So, we need to calculate the intersection, that is, evaluate the system

$$\left\{\begin{array}{c} y=tx-56t+45\\ x^2-y^2=1111\end{array}\right.$$

This is equivalent to

$$x^2-(tx-56t+45)^2=1111$$

or

$$(1-t^2)x^2+(112t^2-90t)x-3136t^2+5040t-3136=0$$

We know x=56 to be a root, so the factoring goes easy enough:

$$(x-56)((1-t^2)x+(56t^2-45t+56))$$

So, we have the following rational solutions:

$$\left(\frac{56t^2-45t+56}{1-t^2},\frac{56t^2-45t+56}{1-t^2}-56t+45\right)$$

In particular, there are infinitely many rational solutions to the equation

$$x^2-y^2=1111$$

and thus there are infinitely many integer solutions to the Diophantine equation

$$a^2-b^2=1111c^2$$

I hope that this was to some interest of people

 

[micromass]

Cool! if you like this type of problem, try the next one/ find a 5-digit number. If we take off the last 3 digits, the new number we obtained is the cube of the original number

You mean to say that the new number is the cube root of the original number??

Anyway, the answer is

Spoiler

32768

 

[xholicwriter]

Thank you for the correction! The answer is correct, but can you show your calculation? I would like to know how you solve it?

 

[micromass]

You want to find a two digit number n such that

$$0<n^3-1000n<1000$$

The first inequality yields that $n>\sqrt{1000}>31$. The second inequality requires us to find an n such that

$$n^3-1000n-1000<0$$

The derivative of this is

$$3n^2-1000$$

and thus the original function has a minimum in (approximatly) 18 and that the function is negative there. After 18, it rises. So the only possible n's are the ones following right after 31. It turns out that 32 was the only answer.

 

[xholicwriter]

You want to find a two digit number n such that

$$0<n^3-1000n<1000$$

The first inequality yields that $n>\sqrt{1000}>31$. The second inequality requires us to find an n such that

$$n^3-1000n-1000<0$$

The derivative of this is

$$3n^2-1000$$

and thus the original function has a minimum in (approximatly) 18 and that the function is negative there. After 18, it rises. So the only possible n's are the ones following right after 31. It turns out that 32 was the only answer.

AWESOME! how can you figure out the solution in such a short period of time? It took me 30' to find out the answer :(

 

[micromass]

AWESOME! how can you figure out the solution in such a short period of time? It took me 30' to find out the answer :(

Sorry
It's experience, I think...

 

[xholicwriter]

Sorry
It's experience, I think...

HAHA, I think so too