Thread by @pgeerkens

@GWOMaths @MathsTim Hint and then solution follows

@GWOMaths @MathsTim Use modulo arithmetic to narrow the possible values of x and y.

@GWOMaths @MathsTim From inspection, x>=5.
As difference of even numbers is always even, x ≅ 1 mod 2.

Evaluating mod 3, noting 18 ≅ 0:
0² ≅ 0; 0².0 ≅ 0
1² ≅ 1; 1².0 ≅ 0
2² ≅ 1; 2².0 ≅ 0
∴ x x ≅ 1 or 2

@GWOMaths @MathsTim Mod 5, noting 18 ≅ 3:
0² ≅ 0; 0².3 ≅ 0
1² ≅ 1; 1².3 ≅ 3
2² ≅ 4; 2².3 ≅ 2
3² ≅ 4; 3².3 ≅ 2
4² ≅ 1; 4².3 ≅ 3
∴
x ≅ 1 or 4 ⇒ y ≅ 0;
x ≅ 2 or 3 ⇒ y ≅ 1 or 4;

Regarding x values 1 .. 25 we have remaining only:
7, 11, 13, 17, 19, 23
w/ restraints on y for each.

@GWOMaths @MathsTim Mod 7 w/ 18 ≅ 4:
0² ≅ 0; 0².4 ≅ 0
1² ≅ 1; 1².4 ≅ 4
2² ≅ 4; 2².4 ≅ 2
3² ≅ 2; 3².4 ≅ 1
4² ≅ 2; 4².4 ≅ 1
5² ≅ 4; 5².4 ≅ 2
6² ≅ 1; 6².4 ≅ 4
∴
x ≅ 1,6 ⇒ y ≅ 0
x ≅ 3,4 ⇒ y ≅ 3,4

∴ x ∈ {11, 17}

@GWOMaths @MathsTim Mod 11 w/ 18 ≅ 7:
0²≅0≅0².7;
1²≅1≅10²; 1².7≅7≅7.10²
2²≅4≅9²; 2².7≅6≅7.9²
3²≅9≅8²; 3².7≅8≅7.8²
4²≅5≅7²; 4².7≅2≅7.7²
5²≅3≅6²; 5².7≅10≅7.6²
∴
x≅0 ⇒ y≅5,6
x≅1,10 ⇒ y≅0
x≅5,6 ⇒ y≅4,7
x≅3,8 ⇒ y≅3,8

∴ x ∈ {11, 17}

@GWOMaths @MathsTim Consider
x = 11
≅ 1 mod 5
≅ 4 mod 7
≅ 0 mod 11
∴ y
< 11
≅ 0 mod 5
≅ 3,4 mod 7
≅ 5,6 mod 11

No such number exists.

∴ x ∈ {17}

@GWOMaths @MathsTim Consider
x = 17
≅ 2 mod 5
≅ 3 mod 7
≅ 6 mod 11

∴ y
< 17
≅ 1,4 mod 5 ⇒ y ∈ {1,4,6,9,11,14}
≅ 3,4 mod 7 ⇒ y ∈ {3,4,10,11}
≅ 4,7 mod 11 ⇒ y ∈ {4,7,15}

Intersection gives ⇒ y ∈ {4}

17² - 18.4² = 289 - 288 = 1

∴ x = 17 and y = 4

@GWOMaths @MathsTim That done, recall that
(x + 1)(x - 1) - x² = 1 for all x

Noting that 18-2 = 16 = 4²

One quickly gets
(18-1)² - 18.4² = 289 - 288 = 1

but without the proof of it being the smallest solution.

@GWOMaths @MathsTim @threadreaderapp unroll