Kvant Math Problem 85

Verified: no
Verdicts: SKIP + SKIP
Solve time: 4m59s
Source on kvant.digital

Problem

Prove that if $a_1$, $a_2$, ..., $a_m$ are pairwise distinct natural numbers, none of which is divisible by the square of an integer greater than one, and $b_1$, $b_2$, ..., $b_m$ are nonzero integers, then $$b_1\sqrt{a_1}+b_2\sqrt{a_2}+\ldots+b_m\sqrt{a_m}\ne 0.$$

L. N. Vasershtein

Exploration

Consider first small examples. If $m=1$, the sum reduces to $b_1\sqrt{a_1}$, which is nonzero since $b_1\ne 0$. For $m=2$, consider distinct square-free natural numbers, such as $a_1=2$, $a_2=3$, with arbitrary nonzero integers $b_1$ and $b_2$. There is no obvious linear combination $b_1\sqrt{2}+b_2\sqrt{3}$ that could vanish. Attempting to falsify this claim, one might seek integers $b_1$ and $b_2$ such that $\sqrt{2}/\sqrt{3}=-b_2/b_1$ is rational, but $\sqrt{2}/\sqrt{3}$ is irrational. This hints at the crucial property: the square roots of distinct square-free numbers are linearly independent over the integers. The main difficulty is proving this linear independence rigorously for any number of terms $m$, not just small examples.

A potential pitfall is assuming irrationality of one ratio suffices for $m>2$, because more than two terms can interact in subtle ways. Another potential trap is neglecting the square-free condition, which is essential for ensuring that no nontrivial integer linear relation exists.

Problem Understanding

The problem asks to prove that if $a_1,\dots,a_m$ are distinct square-free natural numbers and $b_1,\dots,b_m$ are nonzero integers, then the sum $b_1\sqrt{a_1}+\dots+b_m\sqrt{a_m}$ cannot equal zero. This is a Type B problem, a pure proof. The core difficulty is establishing linear independence of square roots of distinct square-free integers over the integers, avoiding any assumptions that rely solely on irrationality of individual terms.

Proof Architecture

Lemma 1: If $\alpha_1,\dots,\alpha_m$ are distinct square-free integers, then the numbers $\sqrt{\alpha_1},\dots,\sqrt{\alpha_m}$ are linearly independent over the rational numbers. Sketch: Suppose a rational linear combination vanishes; clearing denominators produces an integer linear combination that contradicts unique factorization of integers.

Lemma 2: If a nontrivial integer linear combination of square roots of distinct square-free integers is zero, then a corresponding rational combination must also vanish. Sketch: Divide each integer coefficient by their greatest common divisor and treat them as rationals; this reduces to Lemma 1.

The hardest step is Lemma 1, because it requires careful use of prime factorization to exclude any nontrivial relation.

Solution

Assume for contradiction that there exist pairwise distinct square-free natural numbers $a_1,\dots,a_m$ and nonzero integers $b_1,\dots,b_m$ such that

$$b_1\sqrt{a_1} + b_2\sqrt{a_2} + \dots + b_m\sqrt{a_m} = 0.$$

Let $b = \gcd(b_1,\dots,b_m)$ and write $b_i = b c_i$, so that $\gcd(c_1,\dots,c_m) = 1$. Dividing the entire equation by $b$, we obtain an integer linear combination with coefficients $c_i$ satisfying

$$c_1\sqrt{a_1} + c_2\sqrt{a_2} + \dots + c_m\sqrt{a_m} = 0.$$

We will show no such combination exists unless all $c_i = 0$. Proceed by induction on $m$. For $m=1$, $c_1\sqrt{a_1}=0$ implies $c_1=0$, which is impossible. Assume the statement holds for $m-1$. Consider $m$ terms and write

$$c_m \sqrt{a_m} = -\sum_{i=1}^{m-1} c_i \sqrt{a_i}.$$

Squaring both sides, we obtain

$$c_m^2 a_m = \sum_{i=1}^{m-1} c_i^2 a_i + 2\sum_{1\le i<j\le m-1} c_i c_j \sqrt{a_i a_j}.$$

The left-hand side is an integer. On the right-hand side, the terms $c_i^2 a_i$ are integers, and the cross terms $2 c_i c_j \sqrt{a_i a_j}$ are irrational unless $a_i a_j$ is a perfect square. Since $a_i$ and $a_j$ are distinct square-free integers, $a_i a_j$ is never a perfect square, so each $\sqrt{a_i a_j}$ is irrational. The sum of integers plus nonzero irrational numbers cannot vanish. Therefore each $c_i c_j = 0$ for $1\le i<j\le m-1$, implying that at most one $c_i$ is nonzero among $i=1,\dots,m-1$. But then $c_m\sqrt{a_m} = -c_k\sqrt{a_k}$ for some $k$, which is impossible because $\sqrt{a_m}/\sqrt{a_k}$ is irrational. Hence all $c_i = 0$, a contradiction. This completes the induction.

Therefore, no nontrivial integer linear combination of square roots of distinct square-free numbers can vanish. Applying this to the original sum with coefficients $b_i\ne 0$, we conclude that

$$b_1\sqrt{a_1} + \dots + b_m\sqrt{a_m} \ne 0. \quad \Box$$

Verification of Key Steps

The delicate step is squaring the relation and analyzing the cross terms $2 c_i c_j \sqrt{a_i a_j}$. To check rigorously, note that if $a_i$ and $a_j$ are distinct square-free integers, then $a_i a_j$ has no repeated prime factor, hence $\sqrt{a_i a_j}$ is irrational. This is confirmed by factoring $a_i$ and $a_j$ into primes: any repeated prime would contradict the square-free property. Another check is the case $m=2$, where $c_1 \sqrt{a_1} + c_2 \sqrt{a_2} = 0$ implies $\sqrt{a_1}/\sqrt{a_2}=-c_2/c_1$, which cannot be rational because $\sqrt{a_1}/\sqrt{a_2}$ is irrational; this confirms the base of the induction.

Alternative Approaches

An alternative approach uses unique factorization in the ring of integers extended by adjoining $\sqrt{a_1},\dots,\sqrt{a_m}$. One can argue that the norm map from $\mathbb{Z}[\sqrt{a_1},\dots,\sqrt{a_m}]$ to $\mathbb{Z}$ preserves nonvanishing and reduces the problem to checking that no nontrivial integer product of norms can vanish. This method generalizes more naturally to algebraic number fields but requires more background. The main approach is preferable for its elementary use of prime factorization and induction, which is fully accessible to high school students familiar with integers and square roots.