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1. Sorry, the10x10 was wrong.

2. If 2 rooms only touch by the corners, the room which contains the square between those 2 corners will be already have 2 neighbours, and more neighbours will have to touch the rest of that room.

**namealreadychosen**- Replies: 0

Given an n-gon made of meccano, how many extra pieces of meccanno would be needed to brace it so that it cannot flex if they must be placed on the exterior?

I suspect that, if the n-gon is regular and convex the answer is n-1, except when n=3 and no meccano pieces are needed, and that, if it is allowed to be concave, there exists an n-gon for all n which can be braced with n-3 pieces because n-3 pieces will triangulate the interior of any n-gon, and that there exists a convex n-gon for all n>5 which can be braced with

pieces.**namealreadychosen**- Replies: 0

In any magic square of squares, either all the entries are 1(mod3) or all the entries are 0(mod3) because any square is ==0 or 1(mod3) and the sum of each row, column and diagonal, here denoted C, is 3 times the middle entry, therefore C==0(mod3) and only 0,0,0 and 1,1,1 sum to 0(mod3) out of the possible combinations of 0s and 1s. If all the entries are ==0(mod3), they must also be ==0(mod9) because they are square and if each entry is divided by 9 a new magic square of squares will be created. Hence, if there exist any magic squares of squares, there exist some with all entries ==1(mod3) which is ==1 or 4 or 7(mod9).

The only possible combinations summing to 3(mod9) are 1,4,7 and 1,1,1 and 7,7,7 and 4,4,4.

If all entries are the same modulo 9, either they are all identical, which give an infinite number of solutions, or they are different modulo some other power of 3.

Python code to check the C==3(mod9) case

x=[7,4,1]

y=[1,7,4]

z=[7,1,4]

w=[1,4,7]

t=[4,7,1]

u=[4,1,7]

q=[x,y,z,w,t,u]

for a in q:

for b in q:

for c in q:

h=2

for n in range(3):

l=[a[n],b[n],c[n]]

if l not in q:

h=3

if [a[0],b[1],c[2]] not in l or [a[2],b[1],c[0]] not in l:

h=3

if h==2:

print(a,b,c)

print("done")

The only possible combinations summing to 6(mod9) are 1,1,4 and 7,7,1.

Python code to check the C==6(mod9) case

x=[1,1,4]

y=[1,4,1]

z=[4,1,1]

w=[7,7,1]

t=[7,1,7]

u=[1,7,7]

q=[t,u,w,x,y,z]

for a in q:

for b in q:

for c in q:

h=2

for n in range(3):

l=[a[n],b[n],c[n]]

if l not in q:

h=3

if [a[0],b[1],c[2]] not in l or [a[2],b[1],c[0]] not in l:

h=3

if h==2:

print(a,b,c)

print("done")

The only possible combination summing to 0(mod9) are 1,1,7 and 7,7,4 and 4,4,1.

Python code to check the C==0(mod9) case

x=[1,1,7]

y=[7,7,4]

z=[7,1,1]

w=[1,7,1]

t=[4,4,1]

u=[7,4,7]

v=[4,7,7]

r=[4,1,4]

s=[1,4,4]

q=[r,s,t,u,v,w,x,y,z]

for a in q:

for b in q:

for c in q:

h=2

for n in range(3):

l=[a[n],b[n],c[n]]

if l not in q:

h=3

if [a[0],b[1],c[2]] not in l or [a[2],b[1],c[0]] not in l:

h=3

if h==2:

print(a,b,c)

print("done")

My laptop checked through all possible values for the entries (mod9) and I conclude that all magic squares of squares have all entries the same (modulo 9) unless me or the laptop made an error.

18161 = 13.1397

13; 97 = 13.1; 13.7

What if the egg is exactly turned upside down so that every point which was touching the egg before is not touching it any more? It is necessary to include the upper boundary of the egg cup as part of the egg cup for Gurth's theorem to always hold.

This is basically a case of the theorem that every mapping leaves at least one point unchanged.

**namealreadychosen**- Replies: 0

Let m()~n be the number of values which an nth power can take modulo m. e.g. 4()~2=2 because squares can be 0 or 1 modulo 2. I arbitrary invented the symbol ()~ and called it a screw (did I screw up on the name?).

Here is some of what I have worked out:

1()~x=1

2()~x=2

3()~2x-1=3

3()~2x=2

4()~1=4

4()~2x=2

4()~2x+1=3

5()~4x=2

5()~4x-2=3

5()~2x-1=5

6()~2x=4

6()~2x-1=6

7()~1=7

7()~3=4

7()~2x=4

7()~2x+3=7

What is the quickest way to calculate m()~n?

Is n()~n always the smallest factor >1 of n for integers n>1?

When does m()~2=m?

What is the maximum number of values that m()~x can have in terms of m?

Obviously m()~1=m, but are there any simple formulae for m()~n for other positive integers n?

Interesting proofs involving cubes

1. The only integer solution to (x+y+z)^3=xyz is 0,0,0

If, of x,y,z, all are ==1(mod3) or all are ==2(mod3) then x+y+z is a multiple of 3 but xyz cannot be a multiple of 3 if none if its factors is. If all are multiples of 3 then, except in the case where x,y,z=0,0,0, dividing x,y and z by 3 will also yield a solution, so a non-trivial solution with all multiples of 3 only exists if there is a solution with at least one not a multiple of 3.

If, of x,y,z, two are ==1(mod3) and one is ==2(mod3), (x+y+z)^3== (1+1+2)^3==1(mod3) but xyz==2(mod3). If one is ==1(mod3) and 2 are ==2(mod3) then (x+y+z)^3== (1+2+2)^3==2(mod3) but xyz==1(mod3).

2.

since 1=1+1/2 and and (n+2)(n+1)/2=n+1+n(n+1)/2 therefore all successive successors of 1 satisfy this sum (tounge twister ).

since 1^3= ((1+1)/2)^2 and and .

**namealreadychosen**- Replies: 47

How many points can be placed in the plane so that the distance between every pair of points is an integer? What about in higher dimensions?

**namealreadychosen**- Replies: 1

Supposing that there is a set (a,b,c,d,e,f) and a+b, a+c, a+d, b+c... are all squares. Then each member of the set can be paired with 5 others but a+b=b+a, so there are 5*6/2=15 combinations. 5(a+b+c+d+e+f)=sum of all 15 squares.

If a,b and c are odd, a+b, b+c and a+c are even, and must be multiples of 4 since they are square. But a+b+c is odd, and a+b+csad(a+b)+(b+c)+(a+c))/2 which is even. Therefore at most 2 of the integers can be odd.

If a is odd, and the rest are even, a+b,b+c≡1(mod4), b+c≡2-a(mod4), a≡2(mod4).

If a and b are odd, and the rest are even, a+b≡0(mod4), a+c,b+c≡1(mod4), a+2c+b≡2(mod4), c≡1(mod4) but c is meant to be even.

The only remaining possibility is that all 6 are even. Every pair a,b is either both 2(mod4) or both 0(mod4) since (2x)^2==0(mod4). If they are all 0(mod4), they can all be divided by 4 until a set is obtained which is all 2(mod4).

2(mod4)≡2 or 6 (mod8). If a and b are 2(mod8), a+b≡(4(mod8) and cannot be square. If a and b are 6(mod(8), a+b are 12(mod8) and cannot be square. In any set of 6 intergers ≡2(mod8), there will always be a pair both 2(mod8) or both 6(mod8).

Question: What is the largest set possible such that every 3 sum to a square? Are there any sets of 4 such that every 3 sum to a cube?

Also, 23 is the sum of 3 consecutive primes.

If gcd(n,b)>1, there does not exist any sequence of n consecutive primes such that the last digit(s) in base b is n. Does such a sequence always exist when gcd(n,b)=1?

a^2+b^2=1

b/a=tan(n*360/5)

1, 0.31+0.95i, 0.81+0.59i, -0.31-0.95i, -0.81-0.59i

sum=1

an attempt

x=skewes number

x^x^x....x times

iterate x times,

iterate result x times, call answer y,

replace x with y and repeat lines 2-4 y times

call result z

iterate lines 2-5, replacing x with z, z times

call result w

a=largest number expressible with above notation with z symbols

b=largest number expressible with a symbols

repeat finding largest expressible number b times

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