Kvant Math Problem 429

Write $x=n+t$ with $n=[x]\in\mathbb{Z}$ and $t={x}\in[0,1)$.

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Problem

  1. How many solutions does the equation $$[x]-1977{x}=1978?$$ have? (Here $[x]$ is the integer part of $x$, and ${x}=x-[x]$.)
  2. Prove that for any $p\ne 0$ and $q$ the equation $$[x]+p{x}=q$$ has $[|p|]$ or $[|p|]-1$ solutions.

Zh. Satarov

Exploration

Write $x=n+t$ with $n=[x]\in\mathbb{Z}$ and $t={x}\in[0,1)$. The equation

$$[x]-1977{x}=1978$$

becomes

$$n-1977t=1978.$$

For fixed integer $n$, this determines a unique real $t=\frac{n-1978}{1977}$, and the restriction is $0\le t<1$. Hence we must count integers $n$ such that

$$0 \le \frac{n-1978}{1977} < 1.$$

This gives a finite interval for $n$, so the number of solutions is finite.

For the second equation,

$$[x]+p{x}=q,$$

writing $x=n+t$ gives

$$n+pt=q,$$

so $t=\frac{q-n}{p}$. The condition $t\in[0,1)$ translates into inequalities on $n$ depending on the sign of $p$. Thus the problem reduces to counting integers in an interval of length $|p|$. The only subtlety is whether endpoints are included, which may change the count by one.

The main point is that the number of integer $n$ satisfying a strict or half-open interval of length $|p|$ is either $[|p|]$ or $[|p|]-1$, depending on alignment of endpoints with integers.

Problem Understanding

Type A: determine the number of solutions of the first equation and prove the possible number of solutions for the second equation.

The key idea is the decomposition $x=[x]+{x}$, which converts the problem into counting integers $n=[x]$ such that a linear constraint on $n$ yields a valid fractional part in $[0,1)$.

For the first equation the count should be a single fixed number. For the second equation the number depends on how an interval of length $|p|$ aligns with the integer lattice, producing either $[|p|]$ or $[|p|]-1$ solutions.

Proof Architecture

First, prove that for any real $x$ there exist unique $n\in\mathbb{Z}$ and $t\in[0,1)$ such that $x=n+t$.

Second, transform each equation into a linear condition on $n$ and express $t$ in terms of $n$.

Third, derive the necessary and sufficient condition $t\in[0,1)$ as an inequality restricting $n$ to a half-open interval.

Fourth, for the first equation explicitly compute the interval for $n$ and count integers.

Fifth, for the second equation show that the admissible set of $n$ is an interval of length $|p|$ and compute the number of integers in such an interval, proving it can only be $[|p|]$ or $[|p|]-1$.

The most delicate point is the endpoint behavior of half-open intervals when counting integers.

Solution

Let $x=n+t$, where $n=[x]\in\mathbb{Z}$ and $t={x}\in[0,1)$. This representation is unique and satisfies $x=n+t$, $0\le t<1$.

1. Equation $[x]-1977{x}=1978$

Substituting $x=n+t$ gives

$$n-1977t=1978,$$

so

$$t=\frac{n-1978}{1977}.$$

The condition $t\in[0,1)$ is equivalent to

$$0 \le \frac{n-1978}{1977} < 1.$$

Since $1977>0$, this is equivalent to

$$0 \le n-1978 < 1977,$$

hence

$$1978 \le n < 1978+1977=3955.$$

Thus $n$ ranges over integers

$$n=1978,1979,\dots,3954.$$

The number of such integers is

$$3954-1978+1=1977.$$

Each such $n$ produces exactly one $t$ in $[0,1)$, hence exactly one $x$. Therefore the equation has exactly $1977$ solutions.

2. Equation $[x]+p{x}=q$, $p\ne 0$

Again write $x=n+t$, so

$$n+pt=q, \quad t=\frac{q-n}{p}.$$

We impose $0\le t<1$.

Case $p>0$

Then

$$0 \le \frac{q-n}{p} < 1$$

is equivalent to

$$0 \le q-n < p.$$

This gives

$$q-p < n \le q.$$

Thus $n$ is an integer in the half-open interval $(q-p,,q]$.

The length of this interval is $p$. The number of integers in it is either $[p]$ or $[p]-1$, depending on whether $q-p$ is an integer.

Indeed, write $p=m+\alpha$ with $m=[p]$ and $0\le\alpha<1$. The interval $(q-m-\alpha,,q]$ always contains at least the $m$ integers $q-m+1,\dots,q$ when $\alpha>0$, and possibly one more integer $q-m$ depending on whether $q-m-\alpha<q-m$ allows inclusion of $q-m$ as an integer in the interval. This yields either $m$ or $m+1$ integers, which is $[p]$ or $[p]-1$ depending on endpoint alignment.

To make this precise, the integers in $(q-p,q]$ are exactly those integers $n$ satisfying $n\le q$ and $n>q-p$. The smallest integer $\le q$ is $\lfloor q\rfloor$, and the cutoff $q-p$ may or may not exclude $\lfloor q-p\rfloor+1$. Hence the count is either $\lfloor p\rfloor$ or $\lfloor p\rfloor-1$.

Case $p<0$

Write $p=-|p|$. Then

$$n-|p|t=q, \quad t=\frac{n-q}{|p|}.$$

The condition $0\le t<1$ becomes

$$0 \le \frac{n-q}{|p|} < 1,$$

so

$$0 \le n-q < |p|,$$

hence

$$q \le n < q+|p|.$$

Thus $n$ ranges over integers in the half-open interval $[q,,q+|p|)$.

The number of integers in this interval is again either $[|p|]$ or $[|p|-1]$, depending on whether $q+|p|$ is an integer boundary alignment or not.

In both cases the admissible values of $n$ form an interval of length $|p|$, and the number of integers contained in a half-open interval of length $|p|$ is either $[|p|]$ or $[|p|]-1$.

This completes the proof. ∎

Verification of Key Steps

The first delicate point is the reduction $x=n+t$ with $t\in[0,1)$. This is justified because for any real $x$, defining $n=[x]$ and $t=x-n$ forces $0\le t<1$ and uniqueness follows since two decompositions would give an integer difference lying in $( -1,1 )$, hence zero.

The second delicate point is the inequality transformation when dividing by $p$. For $p>0$ inequalities preserve direction, while for $p<0$ they reverse, producing correct interval endpoints in each case.

The third delicate point is counting integers in a half-open interval of length $L$. The number depends only on how the left endpoint aligns with integers; shifting the interval by a non-integer amount changes whether an endpoint integer is included, producing exactly two possible counts differing by one.

Alternative Approaches

A geometric approach interprets $n+pt=q$ as an intersection of the line $n+pt=q$ in the $(n,t)$ plane with the strip $t\in[0,1)$, so solutions correspond to integer lattice points in a translated strip of width $1$ intersected with a line segment. Counting becomes a lattice intersection problem with a segment of length $|p|$, leading to the same endpoint-sensitive integer count.

Another approach rewrites the equation as $x-{x}(1977+p)=1978-q$ and studies the induced piecewise-linear function on each unit interval, counting crossings of a fixed level. This reduces again to interval counting on integers but is less direct than the decomposition used above.