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Here is a very nice (but not easy) elementary problem.
The equality
ceil(2/(2^(1/n)-1)) = floor(2*n/ln(2));

             

is not an identity, it does not hold for each positive integer n.
How to find such a number?

[This is a re-post, because the original vanished when trying a conversion Question-->Post]

The problem appears in the recent book:
Richard P. Stanley - Conversational Problem Solving. AMS, 2020. 

The problem is related to a n-dimensional tic-tac-toe game. The first counterexample (2000) was wrong due to a multiprecision arithmetic error.
The  author of the book writes 
"To my knowledge, only eight values of n are known for which the equation fails,
and it is not known whether there are infinitely many such values",

but using Maple it will be easy to find more.

A brute-force solution is problematic because the smallest counterexample is > 7*10^14.

restart;
a := 2/(2^(1/n)-1): b := 2*n/ln(2):
asympt(b-a, n);

        

It results:  b - a → 1 (for n →oo);
So, to have a counterexample, b must be close to an integer
m ≈ 2*n/ln(2)  ==> n/m ≈ ln(2)/2

The candidates for n/m will be obviously the convergents of the continued fraction of the irrational number ln(2)/2.
 

convert(ln(2)/2, confrac, 200, 'L'):
Digits:=500:
for n in numer~(L[3..]) do
  if not evalf(ceil(a)=floor(b)) then printf("n=%d\n", n) fi;
od:

n=777451915729368
n=140894092055857794
n=1526223088619171207
n=54545811706258836911039145
n=624965662836733496131286135873807507
n=1667672249427111806462471627630318921648499
n=36465374036664559522628534720215805439659141
n=2424113537313479652351566323080535902276508627
n=123447463532804139472316739803506251988903644272
n=97841697218028095572510076719589816668243339678931971
n=5630139432241886550932967438485653485900841911029964871
n=678285039039320287244063811222441860326049085269592368999
n=312248823968901304612135523777926467950572570270886324722782642817828920779530446911
n=5126378297284476009502432193466392279080801593096986305822277185206388903158084832387
n=1868266384496708840693682923003493054768730136715216748598418855972395912786276854715767
n=726011811269636138610858097839553470902342131901683076550627061487326331082639308139922553824778693815

 

So, we have obtained 16 counterexamples. The question whether there are an infinity of such n's remains open.

 

This year, the International Mathematics Competition for University Students  (IMC) took place online (due to Coronavirus), https://www.imc-math.org.uk/?year=2020

One of the sponsors was Maplesoft.


Here is a Maple solution for one of the most difficult problems.

 

Problem 4, Day 1.

A polynomial p with real coeffcients satisfies the equation

p(x+1)-p(x) = x^100, for all real x.

Prove that p(x) <= p(1-x) for   0 <= x and x <= 1/2.

 

A Maple solution.

Obviously, the degree of the polynomial must be 101.

We shall find effectively p(x).

 

restart;

n:=100;

100

(1)

p:= x -> add(a[k]*x^k, k=0..n+1):

collect(expand( p(x+1) - p(x) - x^n ), x):

S:=solve([coeffs(%,x)]):

f:=unapply(expand(eval(p(1-x)-p(x), S)), x);

proc (x) options operator, arrow; (94598037819122125295227433069493721872702841533066936133385696204311395415197247711/16665)*x-37349543370098022593228114650521983084038207650677468129990678687496120882031450*x^3-1185090416633200*x^87+5974737180020*x^89-(86465082200/3)*x^91+133396340*x^93-597520*x^95+2695*x^97-(50/3)*x^99+x^100-(2/101)*x^101+(16293234618989521508515025064456465992824384487957638029599182473343901462949018943/221)*x^5-69298763242215246970576715450882718421982355083931952097853888722419955069286800*x^7+(113991896447569512043394769396957538374962221763587431560580742819193991151970540/3)*x^9-(450021969146981792096716260960657763583495746057337083106755737535521294639081800/33)*x^11+3451079104335626303615205945922095523722898887765464179344409464422173275181060*x^13-648776866983969889704838151840901241863730925272452260127881376737469460326640*x^15+(1224135636503373678241493336115166408006020118605202014423201964267584789018590/13)*x^17-(32609269812588448517851078111423700053874956628293000710950261666057691492700/3)*x^19+(17369174852688147212979009419766100341356836811271344020859968314555332168046/17)*x^21-79714896335448291043424751268405443765709493999285019374276097663327217200*x^23+(26225149723490747954239730131127580683873943002539194987613420614551124468/5)*x^25-294965074792241210541282428184641838437329968596736990461830398732050600*x^27+(186430797065926226062569133543332579493666384095775768758650822594552980/13)*x^29-608766986011732859031810279841713016991034114339196337222615083429200*x^31+22758671683254934243234770245768111655371809025564559292966948184145*x^33-755022138514287934394628273773230341731572817528392747252537299270*x^35+(380420681562789081339436627697748498619486609696130138245054547645/17)*x^37-596110444235534895977389751553577405150617862905657345084592800*x^39+(186546013247587274869312959605954587283787420112828231587660264/13)*x^41-313678397368440441190125909536848768199325715147747522784400*x^43+6254306446857003025144445909566034709396500424382183891144*x^45-114204496639521606716779723226539643746613722246036949600*x^47+1916927215404111401325904884511116319416726263341690260*x^49-29677354167404548158728688629916697559643435320275800*x^51+(93950257927474972838978328999588595121346462082404180/221)*x^53-5650787690628744633775927032927548604440367748960*x^55+69888520126633344286255800412032531913013033640*x^57-806279422358340503473340514496960223283853200*x^59+8696895011389170857678332370276446830499368*x^61-87900576836101226420991143179656778525600*x^63+(10844299000116828980379757772973769420469/13)*x^65-7447304814595165455238549781183862150*x^67+(1065245686771269279784908613651828005/17)*x^69-497741911503981694520541768814800*x^71+3738596479537236832468307626580*x^73-26593490941061853727808593704*x^75+179403449737703736809514420*x^77-1149393958953185579079600*x^79+(21007540356807993839074/3)*x^81-(121855249152521399900/3)*x^83+(3818021878637120462/17)*x^85 end proc

(2)

plot(f, 0..1); # Visual check: f(x)>0 for 0<x<1/2

 

f(0), f(1/4), f(1/2);

0, 2903528346661097497054603834764435875077553006646158945080492319146997643370625023889353447129967354174648294748510553528692457632980625125/3213876088517980551083924184682325205044405987565585670602752, 0

(3)

sturm(f(x), x, 0, 1/2);

1

(4)

So, the polynomial f has a unique zero in the interval (0, 1/2]. Since f(1/2) = 0  and f(1/4) > 0, it results that  f > 0 in the interval  (0, 1/2). Q.E.D.

 

Download imc2020-1-4.mw

Can you guess what P() produces, without executing it?

P:=proc(N:=infinity) local q,r,t,k,n,l,h, f;
q,r,t,k,n,l,h := 1,0,1,1,3,3,0:
while h<N do 
   if 4*q+r-t < n*t
   then f:=`if`(++h mod 50=0,"\n",`if`(h mod 10=0," ","")); printf("%d"||f,n);   
        q,r,t,k,n,l := 10*q,10*(r-n*t),t,k,iquo(10*(3*q+r),t)-10*n,l
   else q,r,t,k,n,l := q*k,(2*q+r)*l,t*l,k+1,iquo(q*(7*k+2)+r*l,t*l),l+2
   fi
od: NULL
end:

I hope you will like it (maybe after execution).

with(plots):
S:=cat("Happy New Year 2020!   "$3):
N:=length(S): a:=0.77*Pi: h:=2*Pi/N:
display(seq(textplot([cos(a-k*h), sin(a-k*h),S[k+1]], 
        rotation=-Pi/2+a-k*h, 'font'=["times","roman",24]), k=0..N-4), axes=none);

 

Maple can easily solve the B4 problem of the Putnam Mathematical Competition 2019  link

 

B4.  Let F be the set of functions f(x,y) that are twice continuously differentiable for x≥1, y≥1 and that satisfy the following two equations:
    x*(diff(f(x, y), x))+y*(diff(f(x, y), y)) = x*y*ln(x*y)

x^2*(diff(f(x, y), x, x))+y^2*(diff(f(x, y), y, y)) = x*y

 

For each f2F, let

 

"m(f) = min[s>=1]  (f(s+1,s+1)-f(s+1,s)-f(s,s+1)+f(s,s))"

 

Determine m(f), and show that it is independent of the choice of f.


 

# Solution

pdsolve({
x*diff(f(x,y),x)+y*diff(f(x,y),y) = x*y*ln(x*y),
x^2*diff(f(x,y),x,x)+y^2*diff(f(x,y),y,y) = x*y
});

{f(x, y) = (1/2)*(x*y+2*_C1)*ln(x*y)-(1/2)*x*y-2*_C1*ln(x)+_C2}

(1)

f:=unapply(rhs(%[]), x,y);

proc (x, y) options operator, arrow; (1/2)*(y*x+2*_C1)*ln(y*x)-(1/2)*y*x-2*_C1*ln(x)+_C2 end proc

(2)

h := f(s+1, s+1) - f(s+1, s) - f(s, s+1) + f(s, s);

(1/2)*((s+1)^2+2*_C1)*ln((s+1)^2)-(1/2)*(s+1)^2-(s*(s+1)+2*_C1)*ln(s*(s+1))+s*(s+1)+(1/2)*(s^2+2*_C1)*ln(s^2)-(1/2)*s^2

(3)

minimize(h, s=1..infinity);

(4+2*_C1)*ln(2)-1/2-(2+2*_C1)*ln(2)

(4)

answer = simplify(%);

answer = 2*ln(2)-1/2

(5)

 


Download putnam2019-b4.mw

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