Hey, this is Presh Talwalkar and you're watching 'Mind Your Decisions'
Here's a really fun problem
Solve 615 plus x squared is equal to 2 to the power of y, over the integers
You're not allowed to use computers or calculators.
I thank Luka Khizambareli from Georgia for suggesting this problem and for sending a solution
Can you figure it out?
Give this problem a try and when you're ready keep watching the video for a solution
So, how can we solve this equation
Well, I'll admit that I did not solve this problem
So I resorted to WolframAlpha
You can solve equations for integer solutions
For example, if you type solve 3x plus 4y is equal to five over the integers Wolfram
Alpha will give you examples of integer solutions. So I thought: since I'm not able to solve this problem
Maybe I can get the result from WolframAlpha
So I typed it in and to my great surprise it said the result is false
The examples of integer solutions it gave were not actual solutions to this equation over the integers
So this is wrong. The problem actually does have the solution
So now let's proceed to solve a problem that WolframAlpha is not yet able to solve
How can we do that?
We'll start with the basic observation. Since x squared is non-negative
2 to the power of y is at least 615
We can then conclude y is greater than 9 because 2 to the power of 9 is equal to 512
We then proceed by considering that if this equation has a solution
Both sides need to have the same last digit
So we'll analyze the very last digit of 615 plus x squared
We'll start out by considering X. The very last digit it could have is from 0 to 9
So then we consider the last digit of x squared. We take the square of each of these digits and consider only the last digit
To get the last digit of 615 + x squared, we add 5 to each of the last digits of x squared.
We now consider the very last digit of 2 to the power of y.
You can figure this out by doing a few calculations. If you calculate the values for y is equal to 1, 2, 3, 4 and so on
You'll see a pattern: the last digit of 2 to the power of y will follow a pattern 2, 4, 8, 6
then the pattern will repeat 2, 4, 8, 6 and so on for every four numbers
Now in order that the last digits match
We look at the last digit of 615 plus x squared and we want to see which ones
correspond to 2, 4, 8 or 6
This only happens when x is equal to 1, 3, 7 or 9 and the last digit will be 6 or 4
So for the digits to match the very last digits x would have to be 1, 3, 7 or 9 and
y would have to be 2, 4, 6, 8 and so on. In other words y would have to be an even number
So we've established that y is even and that'll be the key to solving this problem
Why does this help us?
Well, if y is even that means y is equal to 2 times n, for n an integer
We can then substitute
2 to the (*times) n for y. So we have 615 plus x squared is equal to 2 to the power of 2 n
We then subtract x squared from both sides
And now we use a clever trick
On the right hand side, we have a difference of squares.
So we can use the difference of squares formula
That a squared minus b squared is equal to the quantity a plus b times the quantity a minus b
In this case we can factor that 615 is equal to the quantity
2 to the n plus x times 2 to then minus x
So
We're now going to analyze
615 for two factors: 2 to the n plus x and 2 to the n minus x
We can do this by considering the prime factorization
615 is equal to 3 times 5 times 41
So we consider which two numbers, what pairs of factors we could have which would equal 615
We can enumerate the possibilities
We can have 1 times 615, we can have 3 times 205
5 times 123 or 15 times 41
Now, how can we figure out which possibility corresponds to our solution?
We'll consider adding the two factors together
2 to the n plus x plus 2 to the n minus x will equal
2 times 2 to the n and that simplifies to be 2 to the power of n plus 1
So the sum of the two factors is equal to 2 to the power of n plus 1
So we now consider the sum of the factors in each of these 4 cases
We need it to be a power of 2 and of these 4 numbers:
only 128 is a power of 2
So the sum of factors, if there's a solution, would be 128, which equals 2 to the power of 7
which equals 2 to the power of n plus 1
Therefore n is equal to 6 and then y is equal to 2 times n which is 12
Now with this knowledge, how can we solve for x?
Well, we can consider the difference of factors. 2 to the n plus x minus the quantity 2 to the n minus x is equal to 2x.
And that's the difference of the two factors
So now we take the difference of 123 and 5 and we get 118
That's equal to 2x which means X is equal to 59. So if there's a solution
It would have to be X is equal to 59 and Y is equal to 12 and in fact, we can verify that
615 plus 59 squared is equal to
4096 which exactly equals 2 to the power of 12 and
for completeness
We should also include X is equal to negative 59 and Y is equal to 12 because integers can be both positive and negative
And I'll admit this was a solution I overlooked when I first published this video. Thank you all for catching the mistake
We just solved a problem that WolframAlpha is not yet able to solve
All we did was use the principles of number theories. I think that's pretty neat
Did you figure it out?
Thanks for watching this video
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