MMC 2025 Sol Day 1

Math&Maroc Competition
Solutions day 1
Date: July 26, 2025
Problem 1
The sequence (an)is defined by
a1“1, a2“1, an“öp2an´1`an´2` ¨ ¨ ¨ ` a1ùif n > 2.
Find a2025.
Solution
We will prove, by induction on n, that a2n“a2n`1“nfor every ně1. In particular, we get
that a2025 “1012. The claim is true for n“1.
Assume the claim is true for n. Then
a2n`2“öp2a2n`1`a2n` ¨ ¨ ¨ ` a1ù
“j»2n`n`2((n´1) `(n´2) ` ¨ ¨ ¨ ` 1) `1k
“j»n`n(n`1) `1k“n`1,
and therefore
a2n`3“öp2a2n`2`a2n`1` ¨ ¨ ¨ ` a1ù
“j»2(n`1) `2(n`(n´1) ` ¨ ¨ ¨ ` 1) `1k
“j»(n`2)(n`1) `1k“n`1,
as (n`1)2“n2`2n`1< n2`3n`3< n2`4n`4“(n`2)2.
So the claim is true for every ně1.

Problem 2
For any non-empty subsets Xand Yof Ně1, we define X¨Y:“{xy, x PX, y PY}. For
instance, if
X“{1,2,4}and Y“{3,4,6}
then
X¨Y“{3,4,6,8,12,16,24}.
We say that a subset Sof Ně1is Moroccan if there exists no two subsets Aand Bof Ně1,
each containing at least two elements, such that A¨Band Sare equal.
Prove that the set of perfect powers which are greater or equal to 2025 is Moroccan.
(A perfect power is an integer nk, where n > 1and k > 1are integers.)
Solution
Sol by I.T:
Denote by Pthe set of primes ě2025 and let us take non-empty sets Aand Bsuch that A¨B
is equal to the set of perfect powers ě2025. Our goal is to show that Aor Bis a singleton.
First, notice we must have 1PAYB, otherwise, by taking p‰qPPand using p2, q2PA¨B,
we get p, q PAXB, and so pq PA¨B, which is a contradiction. WLOG, say 1PA. Clearly,
since 1RA¨B, we have 1RB.
Let pPPbe arbitrary, then p2PA¨Byet pRA¨Bso, combined with 1PAand 1RB, we
get that p2PB. This means that all elements of Aare perfect squares (we allow 1). Indeed,
take aPAand pPPnot a divisor of a, then ap2PA¨Bis a perfect power, which can only be
a perfect square, so ais a perfect square.
Let pPPbe arbitrary, then p3PA¨Band Acontaining only perfect squares implies either
p3PBor pPB. The case pPBis incompatible with 1PAand pRA¨B, so p3PB. Similar
to before, we get that all elements of Aare perfect cubes (we allow 1).
We will show by induction that @pPP, @ně2, pnPB. The base cases n“2and n“3were
already proven. Let ně3and assume that @pPP, @kP[2, n], pkPB. Let p‰qPPbe
arbitrary, it is clear that @kP[1, n `1], pkRA. Indeed, assume the opposite and take such a
k. Since Acontains only perfect cubes, kě3. Recall that qk´1PBby induction hypothesis,
as such, pkqk´1PA¨B, which is a contradiction. Now, we know that pn`1PA¨B. The only
possible case is pn`1PB. By induction, we get the result for all ně2.
Finally, using the usual trick on prime values p, we get that Acontains only p-th powers for all
primes p(we allow 1). The only such element is 1, so A“{1}.
Sol by A.K:
Assume the contrary and denote n0“2025. Hence there are sets A, B such that |A|,|B|ě2
and S“A¨B. Let p>n0be any prime – then p2PS, as it is a perfect power. We must have
either pPAXB, or 1PA,p2PB, or p2PA, 1PB.
Suppose, for contradiction, that pPAXB. Take any prime q > n0,q‰p– again, q2PSand
at least one of q2and qmust be in AYB. However, then one of pq, pq2is in S, contradiction.
Hence, for any prime p>n0, we have 1PA,p2PBor p2PA,1PB.
2

Next, we claim it cannot be the case that p2PAand q2PBfor any (not necessarily distinct)
primes p, q. Suppose the contrary. If r > n0, r ‰p, q is a prime, then the perfect power r3is in
S. However, rand r2cannot be in Aand Bin some order, since rp2and rq2are not perfect
powers. But also, 1and r3cannot be in Aor Bin some order, since r3p2and r3q2are not
perfect powers. The claim follows.
From now on, without loss of generality we treat that for any prime p > n0the perfect square
p2is in A, but not in B. Note additionally that Bdoes not contain primes, as if pPB, then
with r2PAfor some prime r > n0, r ‰pwe obtain that pr2is a perfect power, contradiction.
Now, we show by strong induction on kě2that for any prime p > n0the perfect power pkis
in A, but not in B. The base case k“2is covered above. For the induction step, if pkis not
in A, then since none of p, p2, . . . , pk´1are in Bby the induction hypothesis, from the perfect
power pkPSwe obtain pkPB. Take any prime r > n0,r‰p– by the induction hypothesis
we have rk´1PA, which gives a contradiction, as rk´1pkis not a perfect power. Hence pkPA.
By the exact same argument with rwe obtain pkRB. This completes the induction.
To finish off, take a prime p>b, where b > 1is an element of B(possible, since Bhas at least
two elements). For large enough prime q, we have that pqb“anfor some integers a, n. Since
b<p, taking vpgives that q“nvp(a), and since qis prime this means that n“q. Rewrite
the equality as pqb“(px)q, with xan integer, then b“xq, but this is true for infinitely many
primes, contradiction.
3

Problem 3
For pairwise–distinct real numbers a1, . . . , an, prove that:
n
ÿ
j“1
a2
jź
k‰j
aj`ak
aj´ak
“a1` ¨ ¨ ¨ ` an2 .
Solution
1. A convenient rational function. Define
R(x)“x
2
n
ź
k“1
x`ak
x´ak
.(1)
The only poles of Rare at the real numbers x“a1, . . . , an.
2. Partial–fraction decomposition. Because the numerator of Rhas degree one more than
its denominator and that R(x)„
xÑ`8
x
2, there exist constants C, A1, . . . , Ansuch that
R(x)“x
2`C`
n
ÿ
j“1
Aj
x´aj
.(2)
3. Computing each Aj.Multiplying (2) by x´ajand letting xÑajisolates Aj:
Aj“lim
xÑaj
(x´aj)R(x)“aj
2Çlim
xÑaj
(x´aj)
n
ź
k“1
x`ak
x´akå.
Since
lim
xÑaj
(x´aj)
n
ź
k“1
x`ak
x´ak
“2aj
ź
k‰j
aj`ak
aj´ak
,
we obtain
Aj“a2
jź
k‰j
aj`ak
aj´ak
(1 ďjďn).(3)
Define
S:“
n
ÿ
j“1
a2
jź
k‰j
aj`ak
aj´ak
“
n
ÿ
j“1
Aj.(4)
4. Expansion of R(x)at infinity. For any k,
x`ak
x´ak
“1`2ak
x`2a2
k
x2`O
x´3.
Multiplying the nexpansions gives
n
ź
k“1
x`ak
x´ak
“1`2Σ1
x`2Σ2
1
x2`O
x´3,Σ1:“a1` ¨ ¨ ¨ ` an.
4

Hence by (1)
R(x)“x
2`Σ1`Σ2
1
x`O
x´2.(5)
5. Matching the two descriptions. From (2),
R(x)“x
2`C`n
ÿ
j“1
Aj1
x`O
x´2.(6)
Comparing the coefficients of x´1in (5) and (6) yields
n
ÿ
j“1
Aj“Σ2
1.
Using (4),
S“a1` ¨ ¨ ¨ ` an2.
5