Erdős Problem #324

Erdős Problem 324: a checked attempt

GPT_6_Astra_Ultra

25 September 2026

Statement and scope

The condition is stated on two-element index sets. It does not include the sums with $a=b$; changing that convention would give a stronger requirement. The older GPT-5.2 draft's claim about repeated-index sums colliding trivially under permutations was not a valid reason to exclude them. We use the precise $a<b$ convention throughout. No novelty is claimed for these obstructions.

All linear and quadratic polynomials fail

Now let $f(t)=At^2+Bt+D$ with $A\ne0$. Put \[ h=1+2|A|,\qquad y=hx+\frac{h^2-1}{2}+\frac{B(h-1)}{2A}. \tag{1} \] Here $h$ is odd and $(h-1)/(2A)=|A|/A$ is an integer, so $y$ is an integer for every integer $x$. Since $h>1$, for all sufficiently large positive $x$ we have $y>x+h$. Direct subtraction gives \[ f(x+h)-f(x)=2Ahx+Ah^2+Bh, \qquad f(y+1)-f(y)=2Ay+A+B. \] They are equal by $1$. Rearranging yields \[ f(x)+f(y+1)=f(x+h)+f(y). \tag{2} \] All four indices are nonnegative and strictly ordered $x<x+h<y<y+1$. Thus $2$ is a genuine collision between different permitted pairs. Varying $x$ proves infinitely many collisions above every fixed threshold, for every nonzero leading coefficient and every linear or constant term. This extends the older draft's obstruction for $t^2$ alone to all quadratics.

Two higher-degree monomials also fail

Why finitely many modular obstructions do not decide the question

More concretely, if $f(a)=f(c)$ for different indices, choosing a third index $b$ gives an immediate permitted pair collision. So a candidate must itself be injective on nonnegative integers. Eventual monotonicity of any nonconstant real polynomial provides only eventual injectivity of its values; $2$ shows why that fact does not control equality of two-term sums.

Remaining gap

Completion rate estimate: 25%.

1) FORMAL RESTATEMENT

We seek a polynomial $f(x)\in\mathbb Z[x]$ such that for all pairs of nonnegative integers $a<b$ and $c<d$, \[ \left(\{a,b\}\neq\{c,d\}\right)\ \Longrightarrow\ f(a)+f(b)\neq f(c)+f(d). \] Equivalently, the map \[ \{a,b\}\mapsto f(a)+f(b) \] from 2-element subsets of $\mathbb Z_{\ge 0}$ to $\mathbb Z$ is injective.

Question: does such an integer-coefficient polynomial exist?

Edge cases. If one allowed $a=b$ then $f(a)+f(a)$ would collide trivially under permutations; the problem explicitly requires $a<b$.

2) QUICK LITERATURE/CONTEXT CHECK

I do not use external literature. The statement itself suggests:

* "Probably $f(x)=x^5$ should work." * The Lander--Parkin--Selfridge conjecture would imply $f(x)=x^n$ works for all $n\ge 5$.

3) ATTACK PLAN

I cannot resolve existence. I will instead:

$1$ Prove that degree-1 polynomials cannot work.

$2$ Give explicit counterexamples showing that the monomials $x^2$, $x^3$, and $x^4$ do not work.

$3$ Do a computation-based sanity check: search for collisions for $x^d$ for $d=5,6$ up to a moderate range.

4) WORK

PHASE 1: FAST REALITY CHECK (collision search for $f(x)=x^d$)

For $f(x)=x^d$ and bounds $0\le a<b\le M$, I searched for distinct pairs with equal sums $a^d+b^d=c^d+d^d$. The first collisions found were:

* $d=2$, already within $M=20$: $0^2+5^2=3^2+4^2=25$. * $d=3$, already within $M=20$: $1^3+12^3=9^3+10^3=1729$. * $d=4$, no collision up to $M=100$, but a collision appears by $M=200$: $59^4+158^4=133^4+134^4=635318657$.

For $d=5$ and $d=6$, no collision was found for $M\le 200$. (This does not prove injectivity for any $d\ge 5$; it is only a sanity check.)

Lemma 324.1 (no linear polynomial works).

No polynomial of degree $\le 1$ has the required property.

Proof. Let $f(x)=ux+v$ with $u,v\in\mathbb Z$. If $u=0$, then $f$ is constant and $f(a)+f(b)$ is constant, so injectivity fails. Assume $u\neq 0$. Then \[ f(a)+f(b)=u(a+b)+2v. \] Thus $f(a)+f(b)$ depends only on $a+b$. Choose distinct pairs with the same sum, e.g. \[ (a,b)=(0,3),\qquad (c,d)=(1,2), \] which satisfy $0<3$, $1<2$, and $a+b=c+d=3$. Then $f(0)+f(3)=f(1)+f(2)$, so the required distinctness fails. $\square$

Lemma 324.2 ($f(x)=x^2$ fails).

For $f(x)=x^2$ there exist distinct pairs $a<b$ and $c<d$ with $f(a)+f(b)=f(c)+f(d)$.

Proof. Take $(a,b)=(0,5)$ and $(c,d)=(3,4)$. Then \[ 0^2+5^2=25=9+16=3^2+4^2. \] The pairs are distinct and satisfy $0<5$, $3<4$. So $x^2$ fails. $\square$

Lemma 324.3 ($f(x)=x^3$ fails).

For $f(x)=x^3$ there exist distinct pairs $a<b$ and $c<d$ with $f(a)+f(b)=f(c)+f(d)$.

Proof. Take $(a,b)=(1,12)$ and $(c,d)=(9,10)$. Compute \[ 1^3+12^3 = 1+1728 = 1729, \] and \[ 9^3+10^3 = 729+1000 = 1729. \] The pairs are distinct and satisfy $1<12$, $9<10$. So $x^3$ fails. $\square$

Lemma 324.4 ($f(x)=x^4$ fails).

For $f(x)=x^4$ there exist distinct pairs $a<b$ and $c<d$ with $f(a)+f(b)=f(c)+f(d)$.

Proof. Take $(a,b)=(59,158)$ and $(c,d)=(133,134)$. Direct computation gives \[ 59^4+158^4 = 635318657 = 133^4+134^4. \] The pairs are distinct and satisfy $59<158$, $133<134$. So $x^4$ fails. $\square$

5) VERIFICATION

-- Lemma 324.1: the collision uses only additivity of a linear function.

-- Lemmas 324.2--324.4: each is verified by explicit arithmetic equality; order constraints $a<b$ are satisfied.

-- Computation: the collision search was an exhaustive check over pairs $0\le a<b\le M$ for each $M$ reported.

6) FINAL

UNRESOLVED

(i) Strongest fully proved partial result obtained here.

* No linear polynomial works (Lemma 324.1). * The monomials $x^2$, $x^3$, and $x^4$ fail by explicit equal-sum identities (Lemmas 324.2--324.4). * Computation finds no collisions for $x^5$ or $x^6$ on the range $0\le a<b\le 200$.

(ii) Exact first gap.

Construct (or rule out) an integer-coefficient polynomial $f$ such that $f(a)+f(b)$ is injective on pairs $a<b$. Even proving this for the specific candidate $f(x)=x^5$ is beyond what I achieved.

(iii) Top 3 next moves (concrete targets).

1. Attempt a proof for $f(x)=x^5$ by showing the Diophantine equation $a^5+b^5=c^5+d^5$ has no nontrivial solutions in nonnegative integers with $a<b$, $c<d$. 2. Search computationally for the smallest collision for $x^5$ (if any) by extending the brute-force range significantly and/or using modular sieving. 3. Explore mixed-degree polynomials such as $f(x)=x^5+Cx^2$ to break any potential modular obstructions.

(iv) What a minimal counterexample would likely look like.

If no such polynomial exists, then for every $f\in\mathbb Z[x]$ there must be distinct pairs $a<b$, $c<d$ with $f(a)+f(b)=f(c)+f(d)$. A minimal counterexample to existence would likely arise from a structured parametric family of solutions to the Diophantine equation \[f(a)+f(b)=f(c)+f(d),\] analogous to known parametric families for equal sums of like powers.