Method line M15 · When and why do we use it?

Estimation

In the Chinese book: 估计

Integers come one step at a time. Trap the unknown, then check what is left.

📖 §3.4 of 《解题方法与策略》

One idea, one pictureDrag n. Between which two consecutive squares is the middle quantity squeezed? For which n is it not squeezed?

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In one sentence

Real numbers form one continuous line, but integers come one step at a time: there is no integer between and , and no square between and .

So for integers, an inequality is sometimes as strong as an equation:

  • If a quantity is strictly between two consecutive squares, it is not a square.
  • If you can prove that an unknown is at most some number, try the finitely many cases that are left one by one.

“Estimation” means using inequalities to squash infinitely many possibilities down to a few you can count on your fingers.

Predict first

is a positive integer. Can be a perfect square?

Choose an answer first; then you can see the result.

Stop 1 · Common methods for Diophantine equations (Vol. 6)

Change one constant and the story changes: this time there are solutions, and they hide where the squeeze fails.

Example: find all positive integer solutions of y² = x² + x + 7

Step 1 / 6
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  1. It is close to x²

    The right side is a bit bigger than . Is it squeezed between and the next square, ? Try first: , squeezed.

Stop 2 · Integer solutions of equations (Vol. 2)

When there are no “consecutive squares” to squeeze with, try another idea: put the unknowns in order, and the smallest one cannot be big.

Example: positive integers x ≤ y ≤ z satisfy x + y + z = xyz. Find all solutions

Step 1 / 5
  1. A product grows faster than a sum

    When all three numbers are large, is far bigger than . So solutions can only involve small numbers. How do you make “small” precise?

“Assume ” is not laziness: the equation is symmetric in the three letters, and only after putting them in order is there a “largest” one to bound with.

Stop 3 · The Pythagorean equation (Vol. 6)

Integer right triangles also come “one step at a time”: the hypotenuse is longer than the leg , and both are integers, so .

Predict first

A right triangle has positive integer sides, and one leg is . What is the largest possible length of the other leg?

Choose an answer first; then you can see the result.

Example: which integer right triangles have a leg of 9?

Step 1 / 5
  1. Write it as a difference of squares

    Let the other leg be and the hypotenuse .

Stop 4 · Quadratic functions and integer roots (Vol. 3)

For a quadratic equation to have integer roots, the discriminant must be a perfect square. The discriminant is a quadratic in the parameter, just right for squeezing between consecutive squares.

Example: k is a positive integer and x² − kx + k = 0 has integer roots. Find k

Step 1 / 6
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  1. The discriminant must be a square

    The roots are . For them to be integers, must be a perfect square.

All four stops follow the same pattern: first use inequalities to fence off a finite range, then patiently check every case inside it. Neither step can be skipped.

Practice

Exercise 1Basic

The positive integer satisfies . Find .

Exercise 2Advanced

Find all positive integers for which is a perfect square. (Separate them with commas.)

Exercise 3Challenge

Positive integers satisfy . Find all possible values of . (Separate them with commas.)

Mistake Museum · Exhibits from this stop

Before you leave, look at the traps other people fell into here.See every exhibit in the museum →

Exhibit

An inequality true only for large numbers, used for all numbers

The wrong step: Since , lies between two consecutive squares, so it cannot be a perfect square.
What went wrong? Think first, then open

Diagnosis The right-hand inequality simplifies to . For it is simply false: gives and gives , both squares.

The right step: First solve for the range where the inequality holds () and squeeze only there; outside that range, check one by one. The answers are hiding exactly in these small cases.

“Estimation handles the big numbers; check the small ones by hand”

Exhibit

Bounding from one side only

The wrong step: : it is less than a square, so is not a perfect square.
What went wrong? Think first, then open

Diagnosis There are plenty of squares less than : , … A bound on one side rules nothing out. When , is a square.

The right step: Squeeze from both sides, and the two bounds must be consecutive squares: for , , and only that shows it is not a square. Check separately.

“You need both an upper and a lower bound, and they must be consecutive”

Stops on this line

The same idea, used once in each of several places.

  1. Common Methods for Diophantine EquationsVol. 6 Divisibility, Congruences and Diophantine Equations, §3.2
  2. The Pythagorean EquationVol. 6 Divisibility, Congruences and Diophantine Equations, §3.3
  3. Integer Solutions of EquationsVol. 2 Equations and Systems of Equations, ch. 15
  4. Problems on Integer RootsVol. 3 Linear and Quadratic Functions, ch. 6