For a positive integer \(N\) , let \(c_{1}< c_{2}< \dots < c_{m}\) be all the positive integers smaller than \(N\) that are coprime to \(N\) . Find all \(N\geq 3\) such that

\[\gcd (N,c_{i} + c_{i + 1})\neq 1\]

for all \(1\leq i\leq m - 1\)

Here \(\gcd (a,b)\) is the largest positive integer that divides both \(a\) and \(b\) . Integers \(a\) and \(b\) are coprime if \(\gcd (a,b) = 1\) .

Solution:
The answer is all even integers and all powers of 3. First we show that these work.

- When \(N\) is even, all \(c_{i}\) are odd, and so \(2\mid \gcd (N,c_{i} + c_{i + 1})\) for every \(i\)

- When \(N\) is a power of 3, the \(c_{i}\) are exactly the numbers in the range \(1,2,\ldots ,N - 1\) that are not divisible by 3. So, the sequence \(c_{1},\ldots ,c_{m}\) alternates between numbers congruent to 1 (mod 3) and 2 (mod 3). Thus \(3\mid \gcd (N,c_{i} + c_{i + 1})\) for every \(i\) .

Now we show that no other positive integer works. For the sake of contradiction, consider an odd candidate \(N\) that is not a power of 3, and suppose it satisfies the problem condition. Then, since \(c_{1} = 1\) and \(c_{2} = 2\) , we have \(\gcd (N,1 + 2)\neq 1\) , so \(3\mid N\) . Thus, we can write \(N\) as \(3^{k}M\) , where \(k\geq 1\) and \(\gcd (M,6) = 1\) . We have \(M\geq 5\) (as \(M\neq 1\) by assumption) and \(M + 2< N\) .

We now split into two cases based on \(M\) modulo 3.

Case 1. \(M\equiv 1\) (mod 3).

As \(3\mid M - 1\) and \(M\mid M\) , neither \(M - 1\) nor \(M\) are in the sequence \(\{c_{i}\}\) . As \(M\) is odd and \(M\equiv 1\) (mod 3), \(M - 2\) and \(M + 1\) are congruent to 1 (mod 3), are coprime to \(M\) , and are smaller than \(N\) . Thus, both \(M - 2\) and \(M + 1\) are in the sequence, and they are consecutive terms. However, this means \(\gcd (N,2M - 1)\neq 1\) . This gives a contradiction, as both 3 and \(M\) are coprime to \(2M - 1\) .

Case 2. \(M\equiv 2\) (mod 3).

This case is similar to Case 1. Neither \(M\) nor \(M + 1\) are in the sequence, but \(M - 1\) and \(M + 2\) are. We obtain a similar contradiction from \(\gcd (N,2M + 1)\neq 1\) .


Solution:
We give an alternative way to show that any odd \(N\) has to be the power of 3. Suppose for contradiction that \(N\) has at least 2 distinct prime factors. Choose integers \(a,b > 1\) such that

\(a b\mid N\)

- \(a\) and \(b\) are coprime;

- Every prime divisor of \(N\) divides \(a b\) .

We use the Chinese Remainder Theorem, to find an integer \(n\in [1,ab]\) such that

\[\left\{ \begin{array}{l l}{n\equiv 1\pmod {a},}\\ {n\equiv 2\pmod {b}.} \end{array} \right.\]

We claim that there is an index \(i\) such that \(c_{i} = n - 3\) and \(c_{i + 1} = n\) . Indeed, we note the following:

- \(n\) is consecutive to a multiple of \(a\) , so \(\gcd (a, n) = 1\) . In a similar way we get that \(\gcd (b, n - 3) = 1\) .

- \(n\) is 2 away from a multiple of \(b\) , therefore as \(2 \nmid b\) , \(\gcd (b, n) = 1\) . In a similar way we get that \(\gcd (a, n - 3) = 1\) .

- \(n \leq ab \leq N\) and \(n - 3 \geq (a + 1) - 3 > 0\) as \(a > 2\) .

This implies that \(n\) and \(n - 3\) coprime with \(N\) , whereas neither \(n - 2\) nor \(n - 1\) are coprime with \(N\) , which proves our claim. Finally, we claim that \(c_{i} + c_{i + 1} = (n - 3) + n = 2n - 3\) is coprime with \(N\) . Indeed we have \(2n - 3 \equiv - 1 \pmod {a}\) and \(2n - 3 \equiv 1 \pmod {b}\) . This implies that \(N\) is a prime power.

When \(N\) is odd, since \(c_{1} = 1\) and \(c_{2} = 2\) , we have \(\gcd (N, 1 + 2) \neq 1\) , so \(3 \mid N\) , which implies that \(N\) is necessarily a power of 3. This concludes our proof.


An infinite increasing sequence \(a_{1}< a_{2}< a_{3}< \dots\) of positive integers is called central if for every positive integer \(n\) , the arithmetic mean of the first \(a_{n}\) terms of the sequence is equal to \(a_{n}\) .

Show that there exists an infinite sequence \(b_{1},b_{2},b_{3},\ldots\) of positive integers such that for every central sequence \(a_{1},a_{2},a_{3},\ldots\) , there are infinitely many positive integers \(n\) with \(a_{n} = b_{n}\) .

Solution:
We claim that the sequence \(b_{1},b_{2},b_{3},\ldots\) defined by \(b_{i} = 2i - 1\) has this property.

Let \(d_{i} = a_{i} - b_{i} = a_{i} - 2i + 1\) . The condition \(a_{i}< a_{i + 1}\) now becomes \(d_{i} + 2i - 1< d_{i + 1} + 2i + 1\) , which can be rewritten as \(d_{i + 1}\geqslant d_{i} - 1\) . Thus, if \(d_{i + 1}< d_{i}\) , then \(d_{i + 1}\) must be equal to \(d_{i} - 1\) . This implies in particular that if \(d_{i_{0}}\geqslant 0\) but \(d_{i_{1}}\leqslant 0\) for some indices \(i_{1} > i_{0}\) , there must be some intermediate index \(i_{0}\leqslant i\leqslant i_{1}\) with \(d_{i} = 0\) .

Because the average of the first \(a_{n}\) terms of the sequence is equal to \(a_{n}\) , we know for all \(n\) that

\[\sum_{i = 1}^{a_{n}}d_{i} = \sum_{i = 1}^{a_{n}}(a_{i} - 2i + 1) = \sum_{i = 1}^{a_{n}}a_{i} - \sum_{i = 1}^{a_{n}}(2i - 1) = a_{n}^{2} - a_{n}^{2} = 0.\]

Because the sequence \((a_{n})\) is increasing, this implies that the sequence \((d_{i})\) contains infinitely many non- negative \((d_{i}\geqslant 0)\) and infinitely many non- positive \((d_{i}\leqslant 0)\) terms. In particular, we can find arbitrarily large indices \(i_{0}\leqslant i_{1}\) such that \(d_{i_{0}}\geqslant 0\) and \(d_{i_{1}}\leqslant 0\) . By our earlier observation, it follows that there are infinitely many \(i\) such that \(d_{i} = 0\) , as desired.


Solution:
We give an alternative proof that the sequence \(b_{i} = 2i - 1\) works. This proof is by contradiction, so we assume that there are only finitely many \(a_{i}\) such that \(a_{i} = 2i - 1\) .

Let \(S(n) = \sum_{i = 1}^{n}a_{i}\) . We have \(S(a_{n}) = a_{n}^{2}\) and \(S(a_{n + 1}) = a_{n + 1}^{2}\) . If \(a_{n + 1} = a_{n} + 1\) , then it follows that

\[S(a_{n + 1}) - S(a_{n}) = a_{n + 1}^{2} - a_{n}^{2} = a_{n + 1}^{2} - (a_{n + 1} - 1)^{2} = 2a_{n + 1} - 1.\]

On the other hand, if \(a_{n + 1} = a_{n} + 1\) , then \(S(a_{n + 1}) - S(a_{n})\) is \(a_{a_{n + 1}}\) , so it follows that \(a_{a_{n + 1}} = 2a_{n + 1} - 1\) . By assumption, this can only happen finitely many times, so for all sufficiently large \(n\) we must have \(a_{n + 1}\geqslant a_{n} + 2\) .

For large enough \(n\) , we now know that \(a_{n} > 2n - 1\) implies \(a_{n + 1} > (2n - 1) + 2 = 2(n + 1) - 1\) . This means that there are two cases possible:

(A) For all sufficiently large \(n\) (say \(n\geqslant N_{A}\) ) we have \(a_{n} > 2n - 1\)

(B) For all sufficiently large \(n\) (say \(n\geqslant N_{B}\) ) we have \(a_{n}< 2n - 1\)

In case (A), we know for \(m > N_{A}\) that

\[S(m) = S(N_{A}) + \sum_{i = N_{A} + 1}^{m}a_{i}\geqslant S(N_{A}) + \sum_{i = N_{A} + 1}^{n}2i = S(N_{A}) + m(m + 1) - N_{A}(N_{A} + 1)\] \[\qquad = m^{2} + m + S(N_{A}) - N_{A}(N_{A} + 1).\]

For \(m\) large enough (e.g. \(m > N_{A}(N_{A} + 1)\) ), this expression is always larger than \(m^{2}\) , contradicting \(S(a_{n}) = a_{n}^{2}\) for all \(n\) .

Similarly, in case (B), we similarly know for \(m > N_{B}\) that

\[S(m) = S(N_{B}) + \sum_{i = N_{B} + 1}^{m}a_{i}\leqslant S(N_{B}) + \sum_{i = N_{B} + 1}^{n}2(i - 1) = S(N_{B}) + m(m - 1) - N_{B}(N_{B} - 1)\] \[\qquad = m^{2} - m + S(N_{B}) - N_{B}(N_{B} - 1).\]

For \(m\) large enough (e.g. \(m > S(N_{B})\) ), this expression is always smaller than \(m^{2}\) , again contradicting \(S(a_{n}) = a_{n}^{2}\) for all \(n\) .


Solution:
We claim that the sequence \(b_{1}\) , \(b_{2}\) , \(b_{3}\) , ... defined by \(b_{i} = 2i - 1\) has this property.

Lemma. If there are no terms \(a_{j}\) such that \(a_{j} - a_{j - 1} = 1\) , then \(a_{j} = a_{j - 1} + 2\) for all \(j\) .

Proof. Let \(c\) be such that \(a_{d} = c\) for some \(d\) . Now

\[a_{1} + a_{2} + \dots +a_{c} = c^{2}.\]

Equality holds for \(a_{i} = 2i - 1\) for \(1\leq i\leq c\) , so if any difference between two consecutive terms is greater, the left- hand side of the equation is greater than \(c^{2}\) , a contradiction. \(\square\)

Lemma. If both \(d\) and \(d + 1\) are terms of the sequence, i.e. \(a_{c} = d\) and \(a_{c + 1} = d + 1\) for some \(c\) , then \(a_{d + 1} = 2d + 1 = b_{d + 1}\) .

Proof. We have \(a_{1} + a_{2} + \dots +a_{d} = d^{2}\) and \(a_{1} + a_{2} + \dots +a_{d + 1} = (d + 1)^{2}\) . Hence \(a_{d + 1} = (d + 1)^{2} - d^{2} = 2d + 1\) . \(\square\)

From the observations above, we see that we are done if there are infinitely many gaps of size 1. The only remaining case is one with finitely many gaps of size 1. This will be the subject of the following lemma.

Lemma. If there are only finitely many indices \(j\) such that \(a_{j + 1} - a_{j} = 1\) , then there is an index \(n_{0}\) such that for all \(k > n_{0}\) , we have \(a_{k} = 2k - 1\) .

Proof. Let \(r\) and \(s\) be indices such that for all the \(j\) satisfying \(a_{j + 1} - a_{j} = 1\) , we have \(j < r\) , \(s\) . Furthermore, assume \(s > r\) and that there are \(i_{1}\) and \(i_{2}\) such that \(a_{i_{1}} = r\) and \(a_{i_{2}} = s\) . The first goal is to show that \(a_{s} \geq 2s - 1\) . If \(a_{r} \geq 2r - 1\) , this is clearly the case. Assume now \(a_{r} < 2r - 1\) . Now \(a_{r} \geq 2r - 1 - m\) , where \(m\) is the number of indices \(j\) with \(a_{j + 1} - a_{j} = 1\) . Denote \(a_{r + 1} = 2r + 1 - m + \theta_{1}\) , \(a_{r + 2} = 2r + 3 - m + \theta_{2}\) , etc. Remember that \(a_{r + j + 1} - a_{r + j} \geq 2\) always. Now \(0 \leq \theta_{1} \leq \theta_{2} \leq \dots\) . Furthermore, write \(s = r + h\) . Now

\[(r + h)^{2} - r^{2} = a_{r + 1} + a_{r + 2} + \dots +a_{r + h} = \sum_{j = 1}^{h}2r - 1 + 2j - m + \theta_{j}.\]

From this we deduce

\[2r h + h^{2} = 2r h - h - m h + h(h + 1) + \sum_{j = 1}^{h}\theta_{j}.\]

So we obtain \(\sum_{j = 1}^{h}\theta_{j} = m h\) . Since the sequence \(\theta_{j}\) is increasing, we have \(\theta_{h} \geq m\) . Hence, \(a_{s} = a_{r + h} \geq 2r - 1 - m + 2h + m = 2r + 2h - 1 = 2s - 1\) .

Now \(a_{s}\) is exactly the desired shape. If for any \(t > s\) , we have \(a_{t} - a_{t - 1} > 2\) , then

\[a_{s} + a_{s + 1} + \dots +a_{t} > t^{2} - s^{2},\]

again a contradiction.


Solution:
Note that \(a_{1} = 1\) because if it is not the case, then \(a_{1}^{2} = a_{1} + \dots +a_{a_{1}} > a_{1} + a_{1} + \dots +a_{1} = a_{1}^{2}\) .

Assume by contradiction that there are only finitely many indices \(k\) such that \(a_{k} = 2k - 1\) . Set \(i\) to be the largest integer such that \(a_{i} = 2i - 1\) (which must exist as \(a_{1} = 1\) ). Assume that there exists \(j \geqslant i\) such that \(a_{j + 1} - a_{j} = 1\) . Then \(2a_{j + 1} - 1 = a_{j + 1}^{2} - a_{j}^{2} = a_{a_{j + 1}}\) and since \(a_{k} \geqslant k\) for all \(k\) , we have \(a_{j + 1} \geqslant j + 1 > i\) , which contradicts the definition of \(i\) . Thus for all \(j \geqslant i\) , we have \(a_{j + 1} \geqslant a_{j} + 2\) , which implies by induction that \(a_{j} \geqslant 2j - 1\) for \(j \geqslant i\) , and even \(a_{j} \geq 2j\) if \(j > i\) .

There are two ways to finish the solution from here.

## First way to finish the solution

For all \(n\) such that \(a_{n} \geqslant i\) , we have

\[a_{n + 1}^{2} - a_{n}^{2} = a_{a_{n + 1}} + a_{a_{n + 1} - 1} + \dots +a_{a_{n + 1}}\geqslant 2a_{n + 1} + 2(a_{n + 1} - 1) + \dots +2(a_{n} + 1)\] \[\qquad = (a_{n + 1} - a_{n})(a_{n + 1} + a_{n} + 1)\] \[\qquad >a_{n + 1}^{2} - a_{n}^{2}.\]

This gives a contradiction.

## Second way to finish the solution

For all \(n\) such that \(a_{n} \geqslant i\) , we introduce \(x_{n} = a_{n + 1} - a_{n}\) . We have

\[x_{n}^{2} + 2x_{n}a_{n} = a_{n + 1}^{2} - a_{n}^{2} = a_{a_{n + 1}} + a_{a_{n + 1} - 1} + \dots +a_{a_{n} + 1}\geqslant \sum_{j = 1}^{x_{n}}(a_{a_{n}} + 2j)\geq x_{n}a_{a_{n}} + x_{n}(x_{n} + 1).\]

By simplifying, we get \(a_{a_{n}} \leq 2a_{n} - 1\) , which gives a contradiction.

Comment. Proving that \(a_{1} = 1\) is not necessary for this solution. If there exists no \(i\) such that \(a_{i} = 2i - 1\) , then the same argument implies that there exists no \(j\) such that \(a_{j + 1} - a_{j} = 1\) , thus \(a_{j} \geqslant 2j - 1\) for \(j \geqslant 1\) .


Let \(A B C\) be an acute triangle. Points \(B\) , \(D\) , \(E\) , and \(C\) lie on a line in this order and satisfy \(B D =\) \(D E = E C\) . Let \(M\) and \(N\) be the midpoints of \(A D\) and \(A E\) , respectively. Let \(H\) be the orthocentre of triangle \(A D E\) . Let \(P\) and \(Q\) be points on lines \(B M\) and \(C N\) , respectively, such that \(D\) , \(H\) , \(M\) , and \(P\) are concyclic and \(E\) , \(H\) , \(N\) , and \(Q\) are concyclic. Prove that \(P\) , \(Q\) , \(N\) , and \(M\) are concyclic.

The orthocentre of a triangle is the point of intersection of its altitudes.

Solution:
Denote by \(B^{\prime}\) and \(C^{\prime}\) the reflections of \(B\) and \(C\) in \(M\) and \(N\) , respectively. Points \(C^{\prime}\) , \(A\) , \(B^{\prime}\) are clearly collinear and \(D E B^{\prime}A\) is a parallelogram. Since \(E H\perp A D\) , we have \(E H\perp E B^{\prime}\) . Also \(H A\perp A B^{\prime}\) , so points \(H,E,B^{\prime},A\) are concyclic. This gives

\[\angle C^{\prime}Q H = \angle N Q H = \angle N E H = \angle A E H = \angle A B^{\prime}H = \angle C^{\prime}B^{\prime}H,\]

and so points \(C^{\prime}\) , \(B^{\prime}\) , \(Q\) , \(H\) are concyclic. Analogously points \(C^{\prime}\) , \(B^{\prime}\) , \(P\) , \(H\) are concyclic, and so all points \(B^{\prime}\) , \(C^{\prime}\) , \(P\) , \(Q\) , \(H\) are. Now we have

\[\angle N M B^{\prime} = \angle A B^{\prime}M = \angle C^{\prime}B^{\prime}P = \angle C^{\prime}Q P = \angle N Q P,\]

which proves that \(P,Q,N,M\) are also concyclic.

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)


Solution:
Introduce points \(B^{\prime}\) , \(C^{\prime}\) as above. Also define \(A^{\prime} = B^{\prime}E\cap C^{\prime}D\) , so that \(E A D A^{\prime}\) is a parallelogram and \(A D E\) is the medial triangle of \(A^{\prime}B^{\prime}C^{\prime}\) . It that follows that the orthocentre \(H\) of \(A D E\) is the circumcentre of \(A^{\prime}B^{\prime}C^{\prime}\) , and in particular

\[\angle C^{\prime}B^{\prime}H = 90^{\circ} - \angle B^{\prime}A^{\prime}C^{\prime} = 90^{\circ} - \angle D A E = \angle A E H = \angle N E H = \angle N Q H = \angle C^{\prime}Q H.\]

So again we have that \(C^{\prime}\) , \(B^{\prime}\) , \(Q\) , \(H\) are concyclic and conclude as in Solution 1.