Turn a hypothetical solution into a smaller one

Try

Can an integer-sided right triangle have a positive square area? The familiar 3, 4, 5 triangle has area six, and scaling it multiplies area by a square without making six a square. Testing more triples suggests an obstruction but does not prove one. Infinite descent starts by assuming a counterexample exists and then builds a smaller counterexample of exactly the same kind. The contradiction comes from well-ordering, not from an endless computation.

Predict

Choose the hypotenuse as the quantity to minimize among all hypothetical positive integer right triangles with square area. Predict why dividing out a common gcd is safe. If the original area is s² and the side gcd is g, the reduced triangle has area (s/g)², which is an integer. An integer that is a rational square is an integer square: reducing the rational fraction shows its denominator must be one. Therefore a least counterexample can be assumed primitive.

Experiment

Follow the dependency table; each row records a condition needed by the next inference.

Objects	Property	Consequence
Primitive triple	Coprime and one even leg	Parameters m, n
m, n, m+n, m−n	Pairwise coprime positive integers	Square product forces four squares
w, t	Odd with w>t	Half-sum and half-difference are integers
New triangle	Square area and smaller hypotenuse	Contradiction

Try deleting the coprimality condition: two nonsquares such as two and eight can have square product, so that step would fail.

A shorter modular descent: keep the equation unchanged

Before using the right-triangle construction, try a simpler question: can integers x,y,z, not all zero, satisfy x²+y²=3z²? Squares modulo three are zero or one. A sum of two such residues is zero modulo three only when both are zero. Thus three divides x and y. This is a starting implication, not yet the contradiction: nonzero integers can both be divisible by three.

Write x=3X and y=3Y. Substitution gives 3(X²+Y²)=z², so three divides z². Because three is prime, it divides z; write z=3Z. Substitution into the original equation and division by nine gives X²+Y²=3Z². The transformed integers satisfy exactly the same equation.

Now state the well-ordering step. In any nonzero solution, z cannot be zero because a sum of real squares vanishes only when both terms vanish. Replace z by its absolute value and choose a solution with least positive z. The construction produces another solution with positive Z=z/3, strictly less than z. That contradicts the chosen least value. Negative x or y do not affect the argument, and no hypothetical numerical solution was invented.

A modular obstruction needs a smaller integer solution

Equation x²+d y²=m z²m=3, d=1 (prime modulus)m=5, d=2 (prime modulus)m=5, d=1 (prime modulus)m=9, d=1 (composite modulus)Reset descent case

Predict which residues a,b can make a²+d b² divisible by m. Rows label a, columns label b; every cell is the remainder, with zero cells also labelled “zero”.

Remainders modulo 3; all 9 residue pairs

a / b	0	1	2
0	0 (zero)	1	1
1	1	2	2
2	1	2	2

Zero pairs: (0, 0).

Prime modulus and only zero pair: any hypothetical integer solution has both x and y divisible by m. The proof must still derive divisibility of z and verify that division produces the same equation with smaller positive z.

Keep the three proof obligations separate

Exhausting the finite residues proves a congruence implication for every integer representative. It does not alone prove nonexistence of nonzero integer solutions. The general argument additionally proves divisibility of all coordinates and a strictly decreasing positive integer while preserving the equation. No actual solution or simulated descent sequence is claimed in the impossible cases.

Four fixed teaching cases, at most81 cells, temporary state. Retain the full argument and changed-case proof in written practice.

General prime-modulus criterion

Let p be prime and d a positive integer. Suppose the complete residue table has the following property: a²+d b² is zero modulo p only for a=b=0 modulo p. Then x²+d y²=p z² has no integer solution except (0,0,0).

To prove it, a hypothetical nonzero solution has z nonzero because d is positive. Choose least positive |z|. Reducing modulo p and using the table forces p to divide x and y. Write x=pX,y=pY; substitution gives p(X²+dY²)=z². Euclid’s prime-divisor property implies p divides z, since it divides z·z. Write z=pZ. Dividing the original equation by p² gives X²+dY²=pZ², while 0<|Z|=|z|/p<|z|. This is the same forbidden smaller solution. Each step has a role: residue exhaustion gives the first two divisibilities, primality supplies the third, homogeneity preserves the equation, and well-ordering closes the contradiction.

For an odd prime with d not divisible by p, the table hypothesis is equivalent to −d not being a square modulo p. If b is nonzero modulo p, its inverse exists and a²+d b²=0 would imply (a/b)²=−d. If b=0, a²=0 forces a=0. Conversely, a square root r of −d gives the nonzero zero pair (r,1). This field argument explains the table rather than relying on its appearance. The criterion concerns only the indicated equation and hypotheses; a failed criterion alone does not decide whether integer solutions exist.

For p=5,d=2, the squares are zero, one and four, while −2 is three modulo five. Therefore the same descent works. In contrast p=5,d=1 has −1 equal to the square four, and the exact triple (1,2,1) satisfies the equation. A nonzero residue pair alone would not be an integer solution; here substitution supplies a separate existence witness. Replacing a prime by nine also breaks the implication from divisibility of z² to divisibility of z: nine divides 3² but not three. Indeed (3,0,1) solves x²+y²=9z².

Transfer between rational and integer claims

The criterion also rules out rational u,v satisfying u²+d v²=p. If u=A/C and v=B/C with a common positive denominator C, multiplying by C² gives A²+dB²=pC², a nonzero integer triple since C>0. No reduced-fraction assumption is needed; a common denominator suffices. This is a proved reduction, not a numerical search for rational approximations.

Repair two tempting but incomplete arguments

“Both coordinates are divisible by p, so they must be zero” is false. The conclusion follows only after showing all three coordinates divide to another solution with smaller positive |z|. “I checked every triple up to a large bound” also fails to quantify over all integers. A complete finite residue table can support a universal implication because every integer has one of those residues; the subsequent descent then rules out all sizes.

This modular construction addresses the explicitly retained least-counterexample transfer in the Spring2015 MIT6.042J first-midterm audit. These quadratic equations are original changed examples, not a reproduction of that paper’s equation. The earlier right-triangle descent remains a distinct construction; both require proof that the transformed object stays in the same class.

Understand

Write the primitive legs as m²−n² and 2mn, with m>n>0 coprime and opposite parity. Its area is mn(m+n)(m−n). These four factors are pairwise coprime: gcd(m, m±n)=gcd(n, m±n)=1, while a divisor of m+n and m−n divides 2m and 2n and is odd, so it is one. Since their product is a square, each is a square. Write m=u², n=v², m+n=w², m−n=t². The latter two are odd squares, with w>t>0. Put a=(w+t)/2 and b=(w−t)/2. They are positive integers and satisfy a²+b²=(w²+t²)/2=m=u². Also 2ab=(w²−t²)/2=n=v². Because odd squares are one modulo eight, w²−t²=2v² is divisible by eight, so v is even.

Higher roots need denominator and exponent arguments

The cube root of −8/27 is −2/3, while the fourth root of 16/81 is 2/3. But the fourth root of 4/9 is irrational even though its square root is rational. A square-only argument does not classify higher degrees. First reduce the input: 8/18 equals 4/9, so inspecting unreduced numerator exponents would wrongly reject its rational square root.

Which prime exponent blocks a rational root?

Predict the sign restriction, reduce the fraction, then test whether every prime exponent is divisible by the degree. An even root here means the nonnegative real root.

Fraction Rational odd root: -8/27Fourth-power fraction: 16/81Reduce before judging exponents: 8/18Real but irrational higher root: 2/1Zero boundary: 0/25Unit boundary: 1/1Negative unit: -1/1 Degree 2345 Reset

Reduced fraction: -8/27. Rational real root.

Exact root: -2/3. Raising this fraction to power 3 returns the reduced input exactly.

Prime	Exponent in reduced rational	Divisible by degree?
2	3	Yes
3	-3	Yes

Negative exponents represent denominator factors. A failed sign condition means no real root, not an irrational real root. Temporary fixed inputs use exact integer arithmetic; this certificate does not assess an independently written proof.

Fix an integer degree k≥2 and a nonzero rational input A/B in lowest terms, with B>0. If A is negative and k is even, no real number has the required kth power. For positive A choose the nonnegative kth root; for odd k the real root is unique and may be negative. Zero has root zero and is treated separately.

Suppose the chosen root is rational u/v in lowest terms, with v>0. Then uᵏ/vᵏ is also in lowest terms: a prime dividing both would divide both u and v. Uniqueness of a reduced rational representation therefore gives A=uᵏ and B=vᵏ (including the sign of A). Hence every prime exponent in |A| and B is divisible by k. Conversely, if these exponents are divisible by k and the sign permits a real root, divide every exponent by k to construct |u| and v, attach the allowed sign, and verify uᵏ/vᵏ=A/B. This proves necessity and sufficiency. For a positive integer input B=1, rationality forces v=1, so a rational kth root of an integer is an integer. No finite decimal approximation proves that statement.

Equivalently, for a nonzero rational r let vₚ(r) be the numerator exponent minus the denominator exponent. The kth root is rational exactly when all these signed exponents are multiples of k, subject to the real sign condition. Reduction is built into this difference. In the example4/9 the signed exponents are2 and−2: both are divisible by2, but neither by4.

Transfer to rational exponents

Let r>0 be rational and let the rational exponent be h/k in lowest terms, where h is a nonzero integer and k is positive. Define r^(h/k) using the positive kth root. If s=r^(h/k) is rational, then sᵏ=rʰ. Taking signed prime exponents gives k·vₚ(s)=h·vₚ(r). Since gcd(|h|,k)=1, k divides vₚ(r) for every prime. Conversely, if all those exponents are divisible by k, r has a positive rational kth root, and raising that root to integer h (including negative h) remains rational. The denominator of the reduced exponent is essential: r^(2/4) uses denominator2, not4. Exponent zero gives1 for every positive r and is handled separately.

For example32/243=2⁵/3⁵, so its power3/5 equals8/27. Its power2/3 is irrational because the signed exponent5 is not divisible by3. Do not infer rationality of separate radicals from a rational product: the positive cube roots of2 and4 are each irrational by their prime exponents, but their product is2 because its cube is8 and positive cube roots are unique.

Apply

The new integer right triangle has legs a, b and hypotenuse u. Its area is ab/2=v²/4=(v/2)², again a positive square. Its hypotenuse is smaller than the original z=m²+n²=u⁴+n². This contradicts the chosen least hypotenuse. Notice that the new triangle need not already be primitive: it belongs to the same class of square-area integer triangles, which is enough for minimality. Every algebraic construction was conditional on the hypothetical solution; no actual numerical counterexample is needed or possible.

Continue

The theorem also rules out a rational right triangle of area one. Clearing denominators multiplies its area by the square of a positive integer and produces a forbidden integer triangle. This is an example of translating a rational problem into an integer descent. When designing a new descent, name the solution class, the decreasing positive integer, and the proof that the transformed object remains in the class. A smaller real value without a positive integer well-ordering argument is not sufficient.

Return to the number theory unit to choose a targeted retry. Continue with explain a toy rsa round trip.