The pair (a,b) is the same as the pair (b,a)
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y^2 -x^2=35, weer x & y be INTEGER . . . me thank anser=1...36-1=35 . . . next after 36 be 49 & 49-35=14...not square . . . then 64, 64-35=29...not square . . . then 81, 81-35=46...not square . . .
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The answer. I am pretty sure is 1
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Take x and x+1 as an example the difference between their squares is 2x+1. If we put 2x+1=35, then x=17, so 18^2-17^2=324-289=35. What about x and x+2. Here the difference between the squares is an even number so it can't be 35. And x and x+3? The difference is 6x+9. But this time 6 doesn't divide into 26 (35-9). Take the general case of x and x+a. The difference between the squares is 2ax+a^2, so 2ax+a^2=35, and a(2x+a)=35. 35 has two factors 5 and 7. If 2x+a=7 then x=(7-a)/2. If a is 5 then x=1 then 2x+a=7, so x=1 and a=5 is a solution, making the numbers 1 and 6 (36-1=35). So we have two solutions: 17 and 18, and 1 and 6. Note that if we choose the factors 1 and 35 and use a(2x+a)=35, a=1 and x=17, and this is the solution we started with!

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