INMO 2008!
Find p + x + y such that p x = y 4 + 4 where p is a prime and x , y are natural numbers.
The answer is 7.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
2 solutions
Log in to reply
Oh! Sorry it should be y 2 − 2 y + 2 = y 2 + 2 y + 2 − 4 y
What about p=2, x=2 and y=0?
Log in to reply
x and y are positive
y belongs to natural no. We can't take Zero
Just a question, how can we know that y 2 − 2 y + 2 doesn't divide ( y − 1 ) ?
Log in to reply
You can use the remainder theorem and let y = 1 , y 2 − 2 y + 2 = 1 = 0
I like this solution, I didn't do it nearly as rigorously but rather inferred that there was a unique solution based on the requested response then checked for trivial solutions. Perhaps asking for the number of solutions and use a multiple choice answer framework (N=1, 2, finite, infinite) would force people to be more rigorous in their solutions rather than using guess and check or brute force method.
Ans. 7. I wrote a little program in Liberty Basic and checked all prime numbers up to 1013. Here is one solution p=5, x=1, y=1. Because they are asking for p+x+y, the answer is 5+1+1=7. I also found two more examples when p is not a prime number: p=20, x=1, y=2 (ans. 23) and p=85, x=1, y=3 (ans. 89).
We begin with the standard factorization y 4 + 4 = ( y 2 − 2 y + 2 ) ( y 2 + 2 y + 2 ) .
Thus, we have y 2 − 2 y + 2 = p m and y 2 + 2 y + 2 = p n for some positive integers m and n such that m + n = x .
Since y 2 − 2 y + 2 < y 2 + 2 y + 2 , we have m < n so that p m divides p n .
Thus, y 2 − 2 y + 2 divides y 2 + 2 y + 2 .
Writing y 2 − 2 y + 2 = y 2 + 2 y + 2 − 4 y we infer that y 2 − 2 y + 2 divides 4 y and hence y 2 − 2 y + 2 divides 4 y 2 .
But 4 y 2 = 4 ( y 2 − 2 y + 2 ) + 8 ( y − 1 ) thus, y 2 − 2 y + 2 divides 8 ( y − 1 ) , so it also divides 8 .
This gives y 2 − 2 y + 2 = 1 , 2 , 4 , 8 . If y 2 − 2 y + 2 = 1 , then y = 1 And y 4 + 4 = 5 , giving p = 5 and x = 1 .
If y 2 − 2 y + 2 = 2 , then y 2 − 2 y = 0 Giving y = 2 , but then y 4 + 4 = 2 0 is not the power of a prime.
The equations y 2 − 2 y + 2 = 4 and y 2 − 2 y + 2 = 8 have no integer solutions we conclude that ( p , x , y ) = ( 5 , 1 , 1 ) is the only solution, so p + x + y = 7 .