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Croatian Mathematical Competitions

Number theory Difficulty 5.7 AIME, harder Prove it Croatia

Find all integer solutions of the equation x4+68=4y4x^4 + 68 = 4y^4. (Tomislav Pejković)

Solution

Let us rewrite the equation:
x4+68=4y4x^4 + 68 = 4y^4
x4−4y4=−68x^4 - 4y^4 = -68
(x2−2y2)(x2+2y2)=−68 (x^2 - 2y^2)(x^2 + 2y^2) = -68

Now, −68-68 factors as (−1)×68(-1) \times 68, (−2)×34(-2) \times 34, (−4)×17(-4) \times 17, (−17)×4(-17) \times 4, (−34)×2(-34) \times 2, (−68)×1(-68) \times 1 and their negatives. We consider all pairs (a,b)(a, b) such that a×b=−68a \times b = -68.

Let x2−2y2=ax^2 - 2y^2 = a, x2+2y2=bx^2 + 2y^2 = b, so a+b=2x2a + b = 2x^2, b−a=4y2b - a = 4y^2.

Thus,
x2=a+b2 x^2 = \frac{a + b}{2}
y2=b−a4 y^2 = \frac{b - a}{4}

We need x2x^2 and y2y^2 to be integers, so a+ba + b is even and b−ab - a is divisible by 44.

Let us check all possible pairs (a,b)(a, b):

1. a=−1a = -1, b=68b = 68
- a+b=67a + b = 67 (odd), b−a=69b - a = 69 (not divisible by 44)
2. a=−2a = -2, b=34b = 34
- a+b=32a + b = 32 (even), b−a=36b - a = 36 (divisible by 44)
- x2=16x^2 = 16, y2=9y^2 = 9
- x=±4x = \pm 4, y=±3y = \pm 3
3. a=−4a = -4, b=17b = 17
- a+b=13a + b = 13 (odd), b−a=21b - a = 21 (not divisible by 44)
4. a=−17a = -17, b=4b = 4
- a+b=−13a + b = -13 (odd), b−a=21b - a = 21 (not divisible by 44)
5. a=−34a = -34, b=2b = 2
- a+b=−32a + b = -32 (even), b−a=36b - a = 36 (divisible by 44)
- x2=−16x^2 = -16 (not a square), y2=9y^2 = 9
6. a=−68a = -68, b=1b = 1
- a+b=−67a + b = -67 (odd), b−a=69b - a = 69 (not divisible by 44)

Now, the negatives:
7. a=1a = 1, b=−68b = -68
- a+b=−67a + b = -67 (odd), b−a=−69b - a = -69 (not divisible by 44)
8. a=2a = 2, b=−34b = -34
- a+b=−32a + b = -32 (even), b−a=−36b - a = -36 (divisible by 44)
- x2=−16x^2 = -16 (not a square), y2=−9y^2 = -9 (not a square)
9. a=4a = 4, b=−17b = -17
- a+b=−13a + b = -13 (odd), b−a=−21b - a = -21 (not divisible by 44)
10. a=17a = 17, b=−4b = -4
- a+b=13a + b = 13 (odd), b−a=−21b - a = -21 (not divisible by 44)
11. a=34a = 34, b=−2b = -2
- a+b=32a + b = 32 (even), b−a=−36b - a = -36 (divisible by 44)
- x2=16x^2 = 16, y2=−9y^2 = -9 (not a square)
12. a=68a = 68, b=−1b = -1
- a+b=67a + b = 67 (odd), b−a=−69b - a = -69 (not divisible by 44)

So, the only valid case is a=−2a = -2, b=34b = 34:
x2=16  ⟹  x=±4 x^2 = 16 \implies x = \pm 4
y2=9  ⟹  y=±3 y^2 = 9 \implies y = \pm 3

Check in the original equation:
x4+68=4y4 x^4 + 68 = 4y^4
256+68=4×81 256 + 68 = 4 \times 81
324=324 324 = 324

Thus, all integer solutions are:
(x,y)=(4,3),(4,−3),(−4,3),(−4,−3) (x, y) = (4, 3), (4, -3), (-4, 3), (-4, -3)

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