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HMMT November, 2015

Algebra Difficulty 5.0 AIME Prove it United States

Problem:

Consider all functions f:Z→Zf: \mathbb{Z} \rightarrow \mathbb{Z} satisfying
f(f(x)+2x+20)=15 f(f(x)+2x+20)=15
Call an integer nn good if f(n)f(n) can take any integer value. In other words, if we fix nn, for any integer mm, there exists a function ff such that f(n)=mf(n)=m. Find the sum of all good integers xx.

Solution

Solution:

Answer: −35-35

For almost all integers xx, f(x)≠−x−20f(x) \neq -x-20. If f(x)=−x−20f(x) = -x-20, then
f(−x−20+2x+20)=15⟹−x−20=15⟹x=−35 f(-x-20+2x+20) = 15 \Longrightarrow -x-20 = 15 \Longrightarrow x = -35
Now it suffices to prove that the f(−35)f(-35) can take any value.
f(−35)=15f(-35) = 15 in the function f(x)≡15f(x) \equiv 15. Otherwise, set f(−35)=cf(-35) = c, and f(x)=15f(x) = 15 for all other xx. It is easy to check that these functions all work.

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