Find all natural numbers n, such that 7^n + 24^n is a square of integer number

[kaloyankolev]

Hello,

my name is Kaloyan.

I have a problem with the following math problem:
Find all natural numbers n, such that 7^n + 24^n is a square of integer number.

I don't have many ideas but I think that we have to look at two different situations:

1. n is an even number
2. n is an odd number

Then I don't know what to do.

I will be very grateful if someone help me.
Greetings,
Kaloyan

 

[Deleted member 4993]

Hello,

my name is Kaloyan.

I have a problem with the following math problem:
Find all natural numbers n, such that 7^n + 24^n is a square of integer number.

I don't have many ideas but I think that we have to look at two different situations:

1. n is an even number
2. n is an odd number

Then I don't know what to do.

I will be very grateful if someone help me.
Greetings,
Kaloyan

What was the topic (subject) for this HW?

 

[Denis]

Find all natural numbers n, such that 7^n + 24^n is a square of integer number.

Do you mean that the square root of (7^n + 24^n) = integer number;
as example: 7^2 + 24^2 = 49 + 576 = 625 : sqrt(625) = 25 ?

 

[Harry_the_cat]

Do you mean that the square root of (7^n + 24^n) = integer number;
as example: 7^2 + 24^2 = 49 + 576 = 625 : sqrt(625) = 25 ?

Isn't that saying the same thing? I think Kaloyan had it right … 625 is the square of an integer.

My first thought was to consider what digit 7^n + 24^n could end in.

If n = 1, 2, 3, … , the last digit of 7^n cycles through 7, 9, 3, 1, 7, 9, 3, 1, ...

If n = 1, 2, 3, … , the last digit of 24^n cycles through 4, 6, 4, 6, … ​

So, the last digit of the sum 7^n + 24^n cycles through 1, 5, 7, 7, 1, 5, 7, 7, …

Now, no perfect square ends in 7, but a perfect square can end in 1 (if square root ends in 1 or 9) or 5 (if square root ends in 5).

For the sum to end in 1 or 5, n=1, 2 ,not 3, not 4, 5, 6, not7, not 8, etc (but of course not all of these will yield a square as the sum).

… I'm going to lunch, someone else can take over now!

 

[Steven G]

Do you mean that the square root of (7^n + 24^n) = integer number;
as example: 7^2 + 24^2 = 49 + 576 = 625 : sqrt(625) = 25 ?

That does seems to be correct.

 

[kaloyankolev]

What was the topic (subject) for this HW?

I don't understand your question. This is a problem from Mathematics competition.

 

[kaloyankolev]

Thank you!

Isn't that saying the same thing? I think Kaloyan had it right … 625 is the square of an integer.

My first thought was to consider what digit 7^n + 24^n could end in.

If n = 1, 2, 3, … , the last digit of 7^n cycles through 7, 9, 3, 1, 7, 9, 3, 1, ...

If n = 1, 2, 3, … , the last digit of 24^n cycles through 4, 6, 4, 6, … ​

So, the last digit of the sum 7^n + 24^n cycles through 1, 5, 7, 7, 1, 5, 7, 7, …

Now, no perfect square ends in 7, but a perfect square can end in 1 (if square root ends in 1 or 9) or 5 (if square root ends in 5).

For the sum to end in 1 or 5, n=1, 2 ,not 3, not 4, 5, 6, not7, not 8, etc (but of course not all of these will yield a square as the sum).

… I'm going to lunch, someone else can take over now!

Hi,

thank you for the answer. I understand your idea but I don't know what to do after this:
No perfect square ends in 7, but a perfect square can end in 1 (if square root ends in 1 or 9) or 5 (if square root ends in 5).