The only such polynomial is $f(x)=x^{10}+x^9+x^8+x^7+x^6+x^5+1$ Notice that $2^{10}+2^9+2^8+2^7+2^6+2^5+1=2017$, and since all of the coefficients are 1 we can't break this down further and maintain the same sum of coefficients.

hmm.. What about the line between them??? Isn't that a polynomial degree one?

achen29 wrote: hmm.. What about the line between them??? Isn't that a polynomial degree one? It doesn't have nonnegative integer coefficients

lifeisgood03 wrote: The only such polynomial is $f(x)=x^{10}+x^9+x^8+x^7+x^6+x^5+1$ Notice that $2^{10}+2^9+2^8+2^7+2^6+2^5+1=2017$, and since all of the coefficients are 1 we can't break this down further and maintain the same sum of coefficients. Yes. However, I would also mention that to add $2^i$ terms together, you have to make a coefficient at least 1 more than the sum of the coefficients of the terms you combined. But, we already have maxed out our sum of coefficients (from $f(1)=7$) so we can't join to terms together.

lifeisgood03 wrote: The only such polynomial is $f(x)=x^{10}+x^9+x^8+x^7+x^6+x^5+1$ Notice that $2^{10}+2^9+2^8+2^7+2^6+2^5+1=2017$, and since all of the coefficients are 1 we can't break this down further and maintain the same sum of coefficients. but how can you claim so?