arithmetic progression

The Problem:
	Find the positive integers m and n for which the roots of the equations x2−mx+(n+1)=0 and x2−(n+1)x+m=0 are positive integers that, together with m and n , form in some order an arithmetic progression whose sum is 21 .

The solution:

Method 1. Nearly all solvers observed that m+n=10 ; here is their argument. We are given that

r1+s1+r2+s2+m+n=21,

and we know that the roots of a monic quadratic equation add up the the negative of the coefficient of the x -term:

r1+s1=m and r2+s2=n+1.

Put these equations together to get

21=(r1+s1)+(r2+s2)+m+n=m+n+1+m+n=2m+2n+1,

whence m+n=10 , as claimed. Next, because m and n are at most 6 and distinct, the only possibilities are m=6 and n=4 , or m=4 and n=6 . The latter pair produces a pair of quadratics that fail to have integer solutions. With m=6 and n=4 , however, the quadratics become

x2−6x+5=0 and x2−5x+6=0,

whose solutions are 1 and 5 (for the first) and 2 and 3 (for the second), satisfying all our requirements.

Method 2. Here we again use the fact that the sum of the roots of each quadratic equation equals the negative of the coefficient of the x -term and, what's more, their product equals the constant term. Because we are dealing with positive integers that satisfy

r1+s1=m=r2s2, and r2+s2=n+1=r1s1,

we deduce that m and n+1 are at least as large as any of the roots. That means m and n can be only 4, 5, or 6. We know from step 1, moreover, that one of the roots must be 1, and that 1 could not be a root of equation 2 because we would then have 1⋅s2=m , contradicting the requirement that m be distinct from s2 . That leaves only the possibility that 1⋅s2=n+1 , whence s2=5 (it cannot equal 6 because m>s2 ) and therefore n=4 .

Method 3. Here we again use the fact that r1s1=n+1 and r2s2=m . The latter tells us that the product of two of the numbers from 1 to 6 equals the third --- the only possibility is 2⋅3=6 , which gives us right away that m=6 . We are left with 1, 4, and 5 to arrange in the equation r1s1=n+1 , which forces n to be 4.


Method 4. The discriminants of the given quadratics, namely

m2−4(n+1) and (n+1)2−4m,

must be perfect squares. By checking all 30 pairs m and n of distinct integers from 1 to 6, one finds that only the choice n=4 and m=6 produces a perfect square in the second discriminant. Happily, it produces a perfect square also in the first discriminant.

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