[Nitin Prasad]

Consider some polynomial P(x) with degree>1. Observe that for large values of natural numbers n,  $$\frac{4P(n)}{P(P(n))+n}$$ is approximately 0. Using this idea it seems that our polynomial must be a constant polynomial. Though these were loose arguments and now we need to make our arguments precise. For this, few lemmas have been stated without proof.

Lemma 1 : Consider polynomial P(x) ∈ R[x] of degree n>0 such that its leading coefficient is positive. It can be shown that there exist natural number m such that P(x) is strictly increasing function in the interval [m,\infty).

Lemma 2: Consider polynomial P(x) ∈ R[x] of degree n>1 such that its leading coefficient is positive. Then there exist natural number m such that P(x)>m    ∀ x>m and is strictly increasing in the interval [m,\infty)
Proof sketch: Consider polynomial Q(x)=P(x)-x. By lemma 1, there exist m such that Q(x)is strictly increasing in [m,\infty). Hence P(x)-x>0 for all x>m i.e. P(x)>x>m for all x>m.

Solution: We consider following cases for P(x) ∈ R[x]-

1)      Degree of P(x)>1 & leading coefficient is positive: Then by the above lemma 2, there exist natural number m such that P(x) is strictly increasing function in the interval (m,\infty) and P(x)>m for all x>m . Hence we have-

$$P(P(n)-1))+n-1<P⌊P(n)⌋-1+n <⌊P⌊P(n)⌋⌋ + n = 4⌊P(n)⌋<4(P(n)+1)        ∀n ∈ N.$$
$$\Longrightarrow P(P(n)-1))-4P(n)+n-5<0        ∀n ∈ N.$$

Now observe that P(P(n)+1))-4P(n)+n+5  is a polynomial of degree>0 with positive leading coefficient. Hence by lemma1 we arrive at a contradiction.

2)      Degree of P(x)>1 & leading coefficient is negative: Similar to above case and leads to a contradiction

3)      Degree of P(x) at most 1: Consider P(x)=ax+b. Since P(0)=0 Hence b=0. Now we have ⌊a⌊an⌋⌋ + n = 4⌊an ⌋

[Nitin Prasad]

 

 

[Nitin Prasad]

Let R(x)=P(x)-Q(x)

Now we have

$$[Q(x)]+[R(x)]\leq[P(x)]=[Q(x)+R(x)]\leq[Q(x)]+[R(x)]+1$$

Hence we have $$O\leq[R(x)]\leq1$$.

Now every non constant polynomial is unbounded. Hence degree of R(x) cannot be greater than 0.

Therefore degree of R(x) is zero and hence is a constant.

Now if P(x) is a nonconstant polynomial then there exist m such that P(m) is an integer and hence it can be shown that R(x)=O.

If degree of P(x) is 0 then the result follows trivially

[Nitin Prasad]

Let $$\frac{(n+1)^2}{n+7}=m\Longrightarrow n^2+(2-m)n+1-7m$$

 

Now for integral values if m, (discriminant) D must be perfect square. Hence,

$$(2-m)^2-4\times (1-7m)=m^2+24m=(m+12)^2-144$$ is a perfect square

Observe that there are only finitely many pair of perfect squares such that their difference is 144. (Why?)

Look for all those pairs and eventually try to determine m for each case. Now once we know allowed values of m we can always know our required values of n

 

 

[Nitin Prasad]

Let (abc) denote the number 100a+10b+c where a,b,c are digits

Now let (abc), (def), (ghi) be the three 3-digit numbers formed using the digits from 1 to 9.

Now observe that-

$$(abc)+(def)+(ghi)

=100a+10b+c+100d+10e+f+100g+10h+i

=99a+9b+99d+9e+99g+9h+a+b+c+d+e+f+g+h+i$$

Since $$a+b+c+d+e+f+g+h+i= 1+2+3+\cdots +9=45$$

Hence (abc)+(def)+(ghi) is divisible by 9

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