Solving a Cool Diophantine Equation: a^2+a+34=b^2

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PLAYLISTS 🎵 :

[mrnanisissa]

The multiplying-Four Trick never dies.

[ShefsofProblemSolving]

My thinking is actually very different:
Let b=a+k. The equation becomes

34 = (2k-1)a +k*k

If a is positive then k<6 is immediate. Thus checking k=1, 2, 3, 4, 5 quickly finishes the problem.

If a is negative then rewrite the equation as

34 - (2k-1)a = k*k
Now I got stuck here for a tad and then I noticed I could isolate a by writing the whole thing as

a = (k*k - 34)/(2k-1)
Which is a divisibility problem which we can solve for k and forget about the positive and negative cases altogether. We’ll get that 2k-1 divides some constant and finish the problem quickly from there.

[alextang]

By rewrite the equation, we can obtain a^2+2a+1+34-a-1=b^2. ie. (a+1)^2+(33-a)=b^2 So we can easily find that a=33 and b=+/-34 are the solutions. Using the same method, we can also obtain (a+2)^2+(30-3a)=b^2. And then obtain the second pair of solutions are a=10 and b=+/-12. Then using the same method again, we can found the 3rd, 4th solution so on. 😉😉😉😉

[yoav]

It is the first time i see syber gives us some homework, it's about time!😃💯

[olumuyiwaasaolu]

I eventually arrived at the approach given but found the first root by inspection; noting that b² - a² = (b + a)(b - a) = (a + 34)×1. Thus, b = 34, a = 33

[haleshs]

Found a solution (1, 6) almost instantly, but watched the video for the million dollar question: Is it the only solution?

[LifeIsBeautiful-kiky]

put b = a + m/n then solve for a, after finding a make it as integers.

[MichaelRothwell]

This is the solution I found before watching the video - it's the same method that Shefs of Problem Solving came up with. The solution in the video is in fact much simpler (it avoids the more taxing arithmetic used here).

Let r=b-a, so b=a+r.
Then a²+a+34=(a+r)²=a²+2ra+r²
a+34=2ra+r²
34-r²=2ra-a=(2r-1)a
(2r-1)a=34-r²
So 2r-1|34-r² and a=(34-r²)/(2r-1)
So 2r-1|4(34-r²) [here I finally multiplied by 4 as in the video] i.e. 2r-1|136-4r²
Since 2r-1|4r²-1, we have 2r-1|(136-4r²)+(4r²-1) i.e. 2r-1|135.
As 135=5×27=3³×5,
Potential values for 2r-1: ±1, ±3, ±9, ±27, ±5, ±15, ±45, ±135
2r=2, 0, 4, -2, 10, -8, 28, -26, 6, -4, 16, -14, 46, -44, 136, -134
r=1, 0, 2, -1, 5, -4, 14, -13, 3, -2, 8, -7, 23, -22, 68, -67
a=(34-r²)/(2r-1) & b=a+r:
r=1: a=33, b=34
r=0: a=-34, b=-34
r=2: a=10, b=12
r=-1: a=-11, b=-12
r=5: a=1, b=6
r=-4: a=-2, b=-6
r=14: a=-6, b=8
r=-13: a=5, b=-8
r=3: a=5, b=8
r=-2: a=-6, b=-8
r=8: a=-2, b=6
r=-7: a=1, b=-6
r=23: a=-11, b=12
r=-22: a=10, b=-12
r=68: a=-34, b=34
r=-67: a=33, b=-34 (from solution for r=1)

As expected, for each solution (a, b) there is a solution (a, -b).

Summarising, we get:
(a, b) = (-34, ±34), (-11, ±12), (-6, ±8), (-2, ±6), (1, ±6), (5, ±8), (10, ±12), (33, ±34).

[nareshkumar-uonz]

Are there also any other methods to solve this question?

[salviaofficinalis]

This is easy compared to what I have to do.

[williamperez-hernandez]

Are negative answers valid for a Diophantine equation? In his time, there didn't exist the concept of negative numbers.

[kanankazimzada]

Indeed, this solution method is absolutely easy and magnificent