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Solutions to a linear equation and perfect square

  1. Aug 5, 2010 #1
    1) Is there any way to find integer solutions for a linear equation ax+by=c, with a,b,c known and without having an another equation to substitute in the above one.
    ex : 3x+y=394;

    Is there any worst case for c being very large.

    2)Is there any easy and fastest way to get t=115734564=9*(3586^2)
    from t1=116964=9*(12996) to "t" using the equation t=(81*x*x)+(390240*x)+116964

    Actually Here if we dont know the value of "t" and searching for a perfect square using the above equation. "t" is the first perfect square we ecounter at x, and all below x cannot generate a perfect square.

    answer is t=280 , can we get "x" in the shortest way without substituting 1,2,3,...........280 respectively.
    i.e how to get from 12996 to a perfect square using the equation.
    Last edited: Aug 5, 2010
  2. jcsd
  3. Aug 5, 2010 #2


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    Science Advisor

    If you have only the equation ax+ by= c then you can solve for y "in terms of x":
    [tex]y= \frac{c- ax}{b}[/tex]\
    Choose any value at all for x and calculate y from the formula.

    If you also have the requirement that all numbers, a, b, c, and x and y are integers, this is a "Diophantine" equation. It will have no solution if the largest common divisor of a and b does not also divide c. If it does, then you can divide through by that common divisor so we can assume that a and b are relatively prime.

    In the example you give, 3x+ y= 394, note that if x= 1, y= -2, then 3x+ y= 3(1)+ (-2)= 1. Further if we multiply those by 394, so that x= 394, y= -788, then 3x+ y= 3(394)+ (-788)= 394(3(1)+ (-2))= 394(1)= 394.

    But, also, if we take x= 394- k and y= -788+ 3k, then 3x+ y= 3(394- k)+ (-788+ 3k)= 3(394)+ (-788()- 3k+ 3k= 3(394)+ (-788)= 394 again.

    That is, x= 394- k and y= -788+ 3k, for k any integer, are integer solutions to 3x+ y= 394.

    I don't understand what you are asking with "2)Is there any easy and fastest way to get t=115734564=9*(3586^2)
    from t1=116964=9*(12996) to "t" using the equation t=(81*x*x)+(390240*x)+116964"

    In what sense do you get t from t1?
  4. Aug 5, 2010 #3
    u can see t1=116964 and t= (81x^2)+390240x+(t1)
    So we have used t1 in t . i.e by adding something to t1 we get t.

    also t1= 9*12996;
    t = 9*(3568^2);

    so by substituting x=280 in (81x^2)+390240x and adding it to t1 we get 12996 to 3568^2;

    My question is : If we do not know the value of "t" and by substituting the integer values in
    the equation of "t" , to search for a value of "t" which gives a perfect square, we will get it at x=280. So is there any fastest way(method) to say "x" value is 280 without substuting the whole integer values.
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