Problems & Puzzles: Puzzles

 Puzzle 871. Claudio Meller 1476 Claudio Meller, in his always interesting site posted on March 8, 2017, the entry 1476. In this entry he shows a curio property of the pair if integers {184, 345}: 184^1+345^1 is a perfect square, 23^2 184^2+345^2 is a perfect square, 391^2 184^3+345^3 is a perfect square, 6877^2 He made two simple questions about this property: żAre there other primitive pairs as 184 & 345? (it's hard to accept that there is only one primitive solution) żAre there pairs {n1, n2} that produce perfect squares n1^e+n2^e for e=1,2,...m, for m>3? The "primitive" condition stated above, avoid the non-primitive condition of the following infinite set of pairs-solutions: {184*z^2, 345*z^2} for z=2, 3, ... ∞ Up today no one has produced an answer to both questions. Q1. Would you try to respond both questions? By my side, I worked a little bit in the generalization of the puzzle posted by Claudio. My generalization is this one: żAre there set of k integers {n1, n2, ... nk}, k=>2 such that n1^e+n2^e+...nk^e are perfect squares for e=1, 2, ...m, m=>3? Here are just some of my results, for m=3: For k=3, {n1, n2, n3}={108,124,129} and many more solutions For k=4, {n1, n2, n3, n4}={2, 2, 22, 38} and many more solutions For k=4, prime values = {29, 41, 709, 1721}; {977, 1033, 1117, 2957} and for sure many more. But I was unable to find solutions for m>3 with any k=>2 value. Q2. Are solutions for k=>2 and m>3 impossible? Q3. For which k values are possible prime solutions?

Contributions came from Jaroslaw Wroblewski, Seiji Tomita and Jan van Delden

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Jarek wrote:

I have found the second solution

n1=147916017521041
n2=184783370001360

and then searched 147916017521041 in Google. I got:
https://oeis.org/A139265

...

I cannot give a reference, but I believe there is a theorem saying
that x^4+y^4=z^2 has no nontrivial integer solutions.

That means that n1^4+n2^4 is never a square.

...

For m=4, k=3 I believe the problem is hopelessly hard, but I
conjecture that there are no solutions.

...

Regarding k=4, I came up with

n1=p^2
n2=n3=pq
n4=q^2

m=Infinity

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Seiji wrote:

Q1:
X + Y = A^2..................................................(1)
X^2 + Y^2 = B^2..............................................(2)
X^3 + Y^3 = C^2..............................................(3)

Let X= px+1, Y= -px, C= qx+1, then we obtain a solution as x = -(-2q+3p)/(3p^2-q^2).
Then equation (2) becomes to equation (4).
B^2=(8q^2p^2-4q^3p+q^4-12qp^3+9p^4)/((3p^2-q^2)^2)...........(4)
Hence, numerator of equation (4) must be square.
Let U = q/p, then V^2=U^4-4U^3+8U^2-12U+9....................(5)
Transform quartic curve (5) to Weierstrass form (6).
M^2 = N^3-N^2-9N+9...........................................(6)
By using Cremona's mwrank, we obtain the rank=1 and generator is P(-1, -4) on (6).
Hence, elliptic curve (6) has infinitely many rational solutions.
For example, the case of P(-1, -4) on (6), corresponding point is (U,V)=(12/5, -51/25) on (5).
Then we obtain (p,q)=(5, 12) and (X,Y)=(8/23, 15/23).
Accordingly, integer solution is obtained such that (X,Y)=(8/23*23^2, 15/23*23^2)=(184, 345).
Small positive solutions are shown below.
[X , Y]
P    [184, 345]
3P    [184783370001360, 147916017521041]
5p    [221277040503926652746669502425642824, 53160561724398540948189501125090985]

I guess the elliptic curve (6) gives infinitely many positive solutions.

Fermat has already proved that X^4+Y^4=Z^2 has no nontrivial solution.
Hence n1^e+n2^e for e=1,2,...m, for m>3　has no nontrivial solution.

Q2:

Only trivial solution was found using below identity. (a^2)^n+(ab)^n+(ab)^n+(b^2)^n=(a^n+b^n)^2, n is arbitrary.

a^2+ab+ab+b^2=(a+b)^2

(a^2)^2+(ab)^2+(ab)^2+(b^2)^2=(a^2+b^2)^2

(a^2)^3+(ab)^3+(ab)^3+(b^2)^3=(a^3+b^3)^2

(a^2)^4+(ab)^4+(ab)^4+(b^2)^4=(a^4+b^4)^2

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Jan wrote:

Q1 First question:

The given solution is of a special type (A,B)=C with A+B=C^2.

This suggests writing A=ac, B=bc, with c=C:

A+B=C^2          becomes  c(a+b)=C^2

A^2+B^2=D^2 becomes (a^2+b^2)c^2=D^2
A^3+B^3=E^2 becomes (a^3+b^3)c^3=E^2

If we also use that a^3+b^3=(a+b)(a^2+b^2-ab), so create an extra factor c:

a+b=c                        (c=C)
a^2+b^2=d^2          (dc=D)
a^2+b^2-ab=e^2    (ec^2=E)

The second equation has solutions in integers if we choose:

a=m^2-n^2 and b=2mn and d=m^2+n^2

From this we see that it is enough to take (m,n)=1.

The first equation is automatically fullfilled with this choice of a,b.

I searched m in [1..300000] and n in [1..m-1], with (m,n)=1 and found 2 primitive solutions.

The first primitive solution has m=4, n=1 and is given in the question.

The second primitive solution has m=3247, n=1560:

a=  8109409

b=10130640

c= 18240049 (prime)
d= 12976609

e=   9286489

(147916017521041,184783370001360,18240049,236693984013841,3089609202533514140089)

I didn't try to find solutions having a different type, or try to prove they don't exist.

Second question:

There does not exist a solution for the equation X^4+Y^4=Z^2, if XYZ unequal to 0 and X,Y,Z integer.
This was known to Fermat.

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