Kvant Math Problem 905

Consider the equation

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Problem

Prove that the equation $4x^n+(x+1)^2=y^2$ in the natural numbers $x$ and $y$

  1. has no solutions for $n=1$,
  2. has at least two solutions for $n=2$,
  3. has infinitely many solutions for $n=2$,
  4. has no solutions in the natural numbers $n\ge3$.

M. Garaev, 10th-grade student

Exploration

Consider the equation

$4x^n + (x+1)^2 = y^2$

in natural numbers $x$, $y$, and $n$. For small $n$, the equation simplifies, so trying $n=1$ and $n=2$ can give a sense of possible solutions. For $n=1$, the equation becomes $4x + (x+1)^2 = y^2$, which is quadratic in $x$, so factoring or completing the square might reveal integer solutions or their absence. For $n=2$, the equation becomes $4x^2 + (x+1)^2 = y^2$, and small integer values of $x$ produce integer $y$, suggesting at least some solutions exist. Exploring a few small $x$ leads to $(x,y) = (0,1)$, $(1,3)$, $(2,5)$, and so on, hinting at an infinite family. For $n \ge 3$, the left-hand side grows faster than the square of $x$ or $x+1$, so integer solutions may be impossible; examining congruences modulo 4 or modulo small primes may help exclude solutions. The core difficulty lies in proving that for $n\ge3$ no solutions exist and characterizing all solutions for $n=2$.

Problem Understanding

The problem asks to classify the natural number solutions $(x,y,n)$ of the equation $4x^n + (x+1)^2 = y^2$. This is a Type A problem, since it requests a complete description of solutions for each fixed $n$ and claims nonexistence in other cases. The core difficulty is controlling the growth of $4x^n$ relative to $(x+1)^2$ and $y^2$, particularly for $n \ge 3$, and explicitly constructing the infinite family for $n=2$. Intuitively, the equation is quadratic in $y$, so one can attempt factoring the left-hand side as a difference of squares or seek parametrizations that generate integer solutions. For $n=2$, the pattern $y=2x+1$ immediately suggests itself as a potential infinite solution sequence.

Proof Architecture

Lemma 1: For $n=1$, the equation $4x + (x+1)^2 = y^2$ has no solutions in natural numbers. Sketch: Expand $(x+1)^2$, rearrange as $x^2 + 6x + 1 = y^2$, and analyze possible factorizations as consecutive integers or using modular arithmetic.

Lemma 2: For $n=2$, the equation $4x^2 + (x+1)^2 = y^2$ has at least two natural number solutions. Sketch: Expand and simplify to $(2x)^2 + (x+1)^2 = y^2$ and check small integer values of $x$.

Lemma 3: For $n=2$, the equation has infinitely many solutions in natural numbers. Sketch: Observe the pattern $y = 2x+1$ satisfies the equation for all $x$, verify rigorously by substitution.

Lemma 4: For $n \ge 3$, the equation $4x^n + (x+1)^2 = y^2$ has no solutions in natural numbers. Sketch: Compare growth rates, consider factorizations of the left-hand side as a difference of squares, and examine inequalities and parity to exclude solutions.

The hardest part is Lemma 4, because ruling out all possible $x$, $y$ for $n \ge 3$ requires careful estimation and congruence analysis.

Solution

For $n=1$, the equation becomes

$4x + (x+1)^2 = x^2 + 6x + 1 = y^2.$

Rewriting, we have

$x^2 + 6x + 1 = y^2,$

which implies

$(x+3)^2 - y^2 = 8.$

This factors as

$(x+3 - y)(x+3 + y) = 8.$

All factor pairs of 8 in natural numbers are $(1,8)$, $(2,4)$, $(4,2)$, $(8,1)$. Setting $x+3 - y = a$ and $x+3 + y = b$, with $a b = 8$, we solve for $x$ and $y$:

  • $a=1$, $b=8$: $x+3 = \frac{1+8}{2}=4.5$ not integer.
  • $a=2$, $b=4$: $x+3 = 3$, $x=0$, but $0$ is not natural.
  • $a=4$, $b=2$: $x+3 = 3$, $x=0$ again invalid.
  • $a=8$, $b=1$: $x+3=4.5$ not integer.

No factorization yields a natural number solution. Therefore, there are no solutions for $n=1$.

For $n=2$, the equation is

$4x^2 + (x+1)^2 = 4x^2 + x^2 + 2x + 1 = 5x^2 + 2x + 1 = y^2.$

Testing small integers:

  • $x=1$: $5+2+1=8$, not a square.
  • $x=2$: $20+4+1=25$, $y=5$.
  • $x=0$: $0+0+1=1$, $y=1$.

Thus $(x,y)=(0,1)$ and $(2,5)$ are solutions.

Observing the left-hand side as a quadratic in $x$,

$4x^2 + (x+1)^2 = 4x^2 + x^2 + 2x + 1 = 5x^2 + 2x + 1,$

we notice

$(2x+1)^2 = 4x^2 + 4x +1.$

The difference is

$(5x^2 + 2x +1) - (2x+1)^2 = x^2 - 2x = x(x-2).$

Hence $5x^2 + 2x +1 = (2x+1)^2 + x(x-2)$. The extra term $x(x-2)$ is nonnegative for $x \ge 2$, showing the sequence $y=2x+1$ fails for $x>2$ exactly. Another approach is to solve the quadratic Diophantine equation by factoring:

$5x^2 + 2x +1 = y^2 \implies 5x^2 +2x +1 - y^2=0.$

This is a quadratic in $x$ with discriminant

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