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Let $a,b$ be two coprime positive integers. I am asking if $-a^2+2ta-1$ and $-b^2+2tb-1$ are also coprime for any positive integer $t>a$ and $t>b$.

Safwane
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2 Answers2

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There are numerous counter examples even for small values. Just go find one.

Rewrite the expression as $\gcd(2(t-1)a - (a-1)^2 , 2(t-1)b -(b-1)^2)$.

Does this immediately suggest how we could make the GCD (a multiple of) 2?

E.g. $a, b$ odd coprime numbers, $t$ anything, then clearly both terms are even.

Calvin Lin
  • 68,864
  • But it is not true that both terms are even. – Safwane Feb 12 '21 at 15:01
  • @Safwane I agree that it's not true that both terms are even always. Is it ever true that both terms are even? Note that your question is "are these always coprime", so a valid answer is "No, these are not always coprime. here are some instances that they are not coprime". – Calvin Lin Feb 12 '21 at 19:29
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Not a 'real' answer, but it was too big for a comment.

I wrote and ran some Mathematica-code:

In[1]:=ParallelTable[
  If[TrueQ[GCD[a, b] == 1 && 
     GCD[2*t*a - a^2 - 1, 2*t*b - b^2 - 1] == 1 && t > a && 
     t > b], {a, b, t}, Nothing], {a, 2, 10}, {b, 2, 10}, {t, 2, 
   10}] //. {} -> Nothing

Running the code gives:

Out[1]={{{{2, 3, 4}, {2, 3, 6}, {2, 3, 7}, {2, 3, 8}, {2, 3, 9}}, {{2, 5, 
    6}, {2, 5, 7}, {2, 5, 9}, {2, 5, 10}}, {{2, 7, 8}, {2, 7, 
    9}}}, {{{3, 2, 4}, {3, 2, 6}, {3, 2, 7}, {3, 2, 8}, {3, 2, 
    9}}, {{3, 4, 5}, {3, 4, 6}, {3, 4, 7}, {3, 4, 8}, {3, 4, 
    10}}, {{3, 8, 9}}}, {{{4, 3, 5}, {4, 3, 6}, {4, 3, 7}, {4, 3, 
    8}, {4, 3, 10}}, {{4, 5, 6}, {4, 5, 7}, {4, 5, 8}, {4, 5, 9}, {4, 
    5, 10}}, {{4, 7, 8}, {4, 7, 9}}}, {{{5, 2, 6}, {5, 2, 7}, {5, 2, 
    9}, {5, 2, 10}}, {{5, 4, 6}, {5, 4, 7}, {5, 4, 8}, {5, 4, 9}, {5, 
    4, 10}}, {{5, 6, 7}, {5, 6, 8}, {5, 6, 9}, {5, 6, 10}}, {{5, 8, 
    9}, {5, 8, 10}}}, {{{6, 5, 7}, {6, 5, 8}, {6, 5, 9}, {6, 5, 
    10}}, {{6, 7, 8}, {6, 7, 9}, {6, 7, 10}}}, {{{7, 2, 8}, {7, 2, 
    9}}, {{7, 4, 8}, {7, 4, 9}}, {{7, 6, 8}, {7, 6, 9}, {7, 6, 
    10}}, {{7, 8, 9}}}, {{{8, 3, 9}}, {{8, 5, 9}, {8, 5, 10}}, {{8, 7,
     9}}, {{8, 9, 10}}}, {{{9, 8, 10}}}}

So, we can see that you're right!

Jan Eerland
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  • Can you check your code? You used GCD[2*t*a - 1,. It should be $2ta - a^2 - 1$. – Calvin Lin Nov 19 '20 at 19:21
  • @CalvinLin done! – Jan Eerland Nov 20 '20 at 06:38
  • Can you clarify what you mean by "see that you're right"? OP is asking for "for any positive integers ...", whereas your script seems to show that there there are counterexamples. E.g. $a = 3, b = 5, t = $anything doesn't appear, as I stated in my solution. – Calvin Lin Nov 20 '20 at 17:35