New multi-grade solutions $x_1^k+x_2^k+x_3^k=y_1^k+y_2^k+y_3^k=z_1^k+z_2^k+z_3^k, k<5$

Question

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1. What the problem is asking (in plain language)

We are looking for three different triples of integers

[ (x_{1},x_{2},x_{3}),\qquad (y_{1},y_{2},y_{3}),\qquad (z_{1},z_{2},z_{3}) ]

such that the three power‑sums are all the same, namely

[ \begin{aligned} x_{1}^{k}+x_{2}^{k}+x_{3}^{k}&=y_{1}^{k}+y_{2}^{k}+y_{3}^{k}=z_{1}^{k}+z_{2}^{k}+z_{3}^{k}=N
&\qquad\text{for every }k=1,2,3,4 . \end{aligned} ]

(Only the powers (k<5) are required.)
If such a configuration exists we call it a multigrade chain of length three for the exponents (k=1,2,3,4).

The question is: Do any non‑trivial (i.e. not just a permutation of the same three numbers) solutions exist?
If they exist we would like to exhibit at least one, otherwise we would like to prove that none exist.

2. The complete solution

2.1 From power sums to elementary symmetric sums

For a triple ((a,b,c)) define the power sums

[ p_k(a,b,c)=a^{k}+b^{k}+c^{k}, \qquad k=1,2,3,4, ]

and the elementary symmetric sums

[ \begin{aligned} \sigma_1 &=a+b+c,
\sigma_2 &=ab+bc+ca,
\sigma_3 &=abc . \end{aligned} ]

The three elementary symmetric sums are the coefficients (up to sign) of the monic cubic [ \Phi_{a,b,c}(t)= (t-a)(t-b)(t-c)=t^{3}-\sigma_1 t^{2}+\sigma_2 t-\sigma_3 . ]

The link between the power sums and the elementary symmetric sums is given by Newton’s identities:

[ \begin{aligned} p_1 &=\sigma_1,\[2mm] p_2 &=\sigma_1p_1-2\sigma_2,\[2mm] p_3 &=\sigma_1p_2-\sigma_2p_1+3\sigma_3,\[2mm] p_4 &=\sigma_1p_3-\sigma_2p_2+\sigma_3p_1 . \end{aligned} \tag{1} ]

These identities hold for any numbers (a,b,c) (integers, rationals, reals …).

2.2 Equality of the first three power sums forces equality of the elementary sums

Assume we have two triples ((a,b,c)) and ((d,e,f)) such that

[ p_k(a,b,c)=p_k(d,e,f)\quad\text{for }k=1,2,3. \tag{2} ]

Denote by (\sigma_i) the elementary sums of ((a,b,c)) and by (\tau_i) those of ((d,e,f)).

From the first line of (1) we obtain immediately

[ \sigma_1=p_1(a,b,c)=p_1(d,e,f)=\tau_1 . \tag{3} ]

Insert (3) into the second line of (1):

[ p_2=\sigma_1p_1-2\sigma_2 \quad\Longrightarrow\quad 2\sigma_2 = \sigma_1p_1-p_2 . ]

Because the right‑hand side is the same for both triples (by (2) and (3)), we get

[ \sigma_2 = \tau_2 . \tag{4} ]

Now use the third line of (1). With (\sigma_1=\tau_1) and (\sigma_2=\tau_2) we have

[ p_3 = \sigma_1p_2-\sigma_2p_1+3\sigma_3 = \tau_1p_2-\tau_2p_1+3\tau_3 . ]

Since the left‑hand side is the same for the two triples, we obtain

[ 3\sigma_3 = 3\tau_3 \;\Longrightarrow\; \sigma_3 = \tau_3 . \tag{5} ]

Thus equality of the first three power sums forces equality of all three elementary symmetric sums.

2.3 Consequence: the two triples are the same up to order

If (\sigma_i=\tau_i) for (i=1,2,3) then the two cubic polynomials are identical:

[ (t-a)(t-b)(t-c) = t^{3}-\sigma_1t^{2}+\sigma_2t-\sigma_3 = (t-d)(t-e)(t-f). ]

Two monic polynomials are equal iff they have the same set of roots (counted with multiplicity).
Hence the multisets ({a,b,c}) and ({d,e,f}) are equal; the three numbers of one triple are just a permutation of the three numbers of the other triple.

In particular, any non‑trivial triple ((a,b,c)) cannot have a distinct triple ((d,e,f)) with the same power sums for (k=1,2,3).

2.4 Adding the fourth power – no new information

If the power sums are already equal for (k=1,2,3), the fourth identity in (1) shows that the fourth power sums are automatically equal:

[ p_4 = \sigma_1p_3-\sigma_2p_2+\sigma_3p_1 ]

and the right‑hand side depends only on (\sigma_1,\sigma_2,\sigma_3) and on the already equal power sums (p_1,p_2,p_3).
Therefore the condition for (k=4) does not introduce any extra restriction; it is a consequence of the first three.

2.5 Extending to three triples

Suppose we have three triples ((x_1,x_2,x_3), (y_1,y_2,y_3), (z_1,z_2,z_3)) satisfying

[ p_k(x)=p_k(y)=p_k(z)\qquad\text{for }k=1,2,3,4 . \tag{6} ]

Apply the result of §2.3 to the pair ((x)) and ((y)): they must be the same multiset. Apply the same argument to the pair ((x)) and ((z)): they also must be the same multiset. Consequently all three triples consist of the same three numbers, perhaps in a different order.

Thus there is no way to obtain three different triples with the same power‑sums for the exponents (1,2,3,4).

2.6 Final statement

Theorem.
Let ((a,b,c),(d,e,f),(g,h,i)) be integer (or rational) triples.
If for every (k=1,2,3,4) [ a^{k}+b^{k}+c^{k}=d^{k}+e^{k}+f^{k}=g^{k}+h^{k}+i^{k}, ] then the three triples are identical up to permutation; i.e.
({a,b,c}={d,e,f}={g,h,i}).

Therefore non‑trivial multigrade chains of length three do not exist for the exponents (k<5). The red cells in the table of the original post must stay empty.

3. Final answer

No non‑trivial integer (or rational) solutions exist for

[ x_1^k+x_2^k+x_3^k = y_1^k+y_2^k+y_3^k = z_1^k+z_2^k+z_3^k\quad\text{for all }k=1,2,3,4 . ]

If the equalities hold for the three exponents (k=1,2,3) (and consequently for (k=4)), the three triples must be the same set of numbers, possibly reordered. Hence a “multigrade chain’’ with three distinct triples cannot be constructed for any (k<5).

4. Common mistakes for this type of problem

Mistake	Why it is wrong	How to avoid it
Assuming the four equations are independent.	The fourth power‑sum is a consequence of the first three via Newton’s identities; it adds no new condition.	Write down Newton’s identities and check the dependence before trying to solve the system.
Treating the three power‑sum equations as linear in the unknown numbers.	Power sums are highly non‑linear; linear methods (e.g., Gaussian elimination) are not applicable.	Use symmetric‑function theory (Newton’s identities) to translate the problem into statements about elementary symmetric sums.
Searching for solutions by brute force without using the symmetry.	The search space grows explosively (≈ (N^{3}) for numbers up to size (N)).	First prove that a non‑trivial solution cannot exist; then a brute‑force search is unnecessary.
Confusing “different triples’’ with “different orderings’’	Permuting the entries of a triple does not give a new solution; it is the same multiset.	After obtaining equal elementary symmetric sums, conclude that the sets of numbers are identical, regardless of order.
Neglecting the possibility of repeated numbers (e.g., (a=b)).	Repeated entries still satisfy the same symmetric‑function relations; the proof works for any multiplicities.	Keep the proof general; no assumption about distinctness is needed.

By keeping these points in mind, one can avoid dead‑ends and reach the decisive conclusion that multigrade chains of length three for (k=1,2,3,4) are impossible.

Original question: New multi-grade solutions $x_1^k+x_2^k+x_3^k=y_1^k+y_2^k+y_3^k=z_1^k+z_2^k+z_3^k, k<5$ on Mathematics Stack Exchange, licensed CC BY-SA.