An indefinite quadratic Diophantine system that turns out to be finite

gengarHours·Dec 17, 2025(edited)

The first equation has a negative square, so I expected infinitely many solutions. How does the linear equation cut the possibilities down? Problem: Find all combinations of four integers

(a, b, c, d)

satisfying the equations

\begin{cases} -a^2 + b^2 + c^2 + d^2 = 1 \\ 3a + b + c + d = 1. \end{cases}

Source: Baltic Way 2016 · MNET-01CT

Open MNET-01CT in the question bank →

3 replies

thecakewaslag·Dec 17, 2025(edited)

Apply Cauchy to the other three variables:

(1-3a)^2=(b+c+d)^2\le3(b^2+c^2+d^2)=3(1+a^2)

. Thus

3a^2-3a-1\le0

gengarHours·Dec 17, 2025(edited)

The roots are

(3\pm\sqrt{21})/6

, so the only integer possibilities are a=0 and a=1. I noticed a different interval in the stored solution; the quadratic itself is enough to avoid that typo.

thecakewaslag·Dec 18, 2025(edited)

For a=0, the other squares sum to 1 and their values sum to 1: (b,c,d) is a permutation of (1,0,0). For a=1, the squares sum to 2 and the values sum to -2: a permutation of (-1,-1,0). All six quadruples check in both equations.

Join the discussion

A hint, a different approach, or a thoughtful question can help.

Loading discussion…

An indefinite quadratic Diophantine system that turns out to be finite

gengarHours·Dec 17, 2025(edited)

The first equation has a negative square, so I expected infinitely many solutions. How does the linear equation cut the possibilities down? Problem: Find all combinations of four integers

(a, b, c, d)

satisfying the equations

\begin{cases} -a^2 + b^2 + c^2 + d^2 = 1 \\ 3a + b + c + d = 1. \end{cases}

Source: Baltic Way 2016 · MNET-01CT

Open MNET-01CT in the question bank →

3 replies

thecakewaslag·Dec 17, 2025(edited)

Apply Cauchy to the other three variables:

(1-3a)^2=(b+c+d)^2\le3(b^2+c^2+d^2)=3(1+a^2)

. Thus

3a^2-3a-1\le0

gengarHours·Dec 17, 2025(edited)

The roots are

(3\pm\sqrt{21})/6

, so the only integer possibilities are a=0 and a=1. I noticed a different interval in the stored solution; the quadratic itself is enough to avoid that typo.

thecakewaslag·Dec 18, 2025(edited)