Suppose that m and n are integers, such

tibukooinm

tibukooinm

Answered question

2022-04-01

Suppose that m and n are integers, such that both the quadratic equations
x2+mxn=0
and x2mx+n=0
have integer roots. Prove that n is divisible by 6.

Answer & Explanation

kattylouxlvc

kattylouxlvc

Beginner2022-04-02Added 11 answers

Now as you got m2+4n=p2 and m24n=q2, we solve further by taking cases.
Therefore let m=2k for some positive integer k and by judging the equation we can see that p and q are even too. Let p=2a and q=2b for some positive integers a and b. Substituting the values of m,p and q, we get
k2+n=a2

k2n=b2
This implies a2+b2=2k2

a2b2=2n
Now let's assume that n is odd, but that would mean a2b2 is not divisible by 4, so either a21±mod4 and b20±od{4} r vice versa (Remember that a square is always 0~or~1±mod4.
Therefore a2+b21±mod4 in both cases. But we have on the other side a2+b2=2k2 which is always either 0±mod4 or 2±mod4.
Thus, we get a contradiction. This implies n is even.
Now we assume n is not divisible by 3 i.e either 2n1±mod3 or 2n2±mod3. Now a square is always either 0~or~1±mod3.
Therefore a2b2=2n2±mod3 is never possible and thus the remaining possibility is a2b2=2n1±mod3. This implies a21±mod3 and b20±mod3. Therefore, a2+b21±mod3, but we had from the other equation a2+b2=2k2 which is always 0~or~2±mod3. Thus, we get a contradiction. Hence, n is divisible by 3 as well. Thus, n is divisible by 6.
Similar analysis goes for the other case where m is odd.

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