Perfect Trinomial Square

We require that $a^{2} + 3ab + b^{2} = n^{2}$ for some positive integer $n$ . Rewriting this equation as

$a^{2} + 3b*a + (b^{2} - n^{2}) = 0$ , we see that

$a = \dfrac{-3b \pm \sqrt{9b^{2} - 4(b^{2} - n^{2})}}{2} = \dfrac{-3b \pm \sqrt{5b^{2} + 4n^{2}}}{2}$ .

Now since we want $a$ to be a positive integer we will take the positive root and require that

$5b^{2} + 4n^{2} = m^{2}$ for some positive integer $m$ . This last equation can then be written as

$5b^{2} = m^{2} - (2n)^{2} = (m - 2n)(m + 2n).$

Now as $5$ is prime it must divide at least one of $m - 2n$ or $m + 2n$ , and since we are looking for the smallest possible value of $a + b$ we will want to minimize $m$ and $n$ as well. With this in mind, we can try setting $m - 2n = 5$ and $m + 2n = b^{2}$ and add these two equations, which yields $2m = b^{2} + 5$ . Since $2m$ is even we will require that $b$ is odd.

Checking successive odd integers, we obtain the following results:

• $b = 1 \Longrightarrow m = 3 \Longrightarrow a = 0$ : discard

• $b = 3 \Longrightarrow m = 7 \Longrightarrow a = -1$ : discard

• $b = 5 \Longrightarrow m = 15 \Longrightarrow a = 0$ : discard

• $b = 7 \Longrightarrow m = 27 \Longrightarrow a = 3 \Longrightarrow a + b = 10$

• $b = 9 \Longrightarrow m = 43 \Longrightarrow a = 8 \Longrightarrow a + b = 17$ .

Any larger values of $b$ will necessarily result in a value for $a + b$ in excess of $10$ , so the minimum possible value for $a + b$ under the given conditions is $\boxed{10}$ .

The algorithm for computer or calculator is as under.
$\text{Set m=10, n=10, p=1.}\\ \text{ These values have to be increased if the answer is not obtained with in this limit.} \\ \text{**........outer loop, b=p add 1 upto n.}\\ \text{...................inner loop, a=1 add 1 upto m.}\\ .........................x=\sqrt{a^2+3ab+b^2}\\ \text{...............................if x integer Print.}\\ \text{..................Next inner loop.}\\ \text{..........Next outer loop.}\\ \text{....... p=n: n=n+10 : End if n>100 Go to **}\\ \text{I have used this with TI-83 and got answer with a=3, and b=7.} \\ \text{I went manually through 37 loop, changing a and b after a=10, to a=0 and b=b+1. }\\ \text{My expression was to start, b=1, a=0.}\\ a+ ~1 ~STR ~ a ~~{\Huge :}\sqrt{ a^2 +3ab+ b^2}\text{ stopped looping when saw the integer answer.}$ .

${ a }^{ 2 }+2ab+{ b }^{ 2 }={ (a+b) }^{ 2 }$

We let ${ a }^{ 2 }+3ab+{ b }^{ 2 }={ x }^{ 2 }$ and x is an integer.

Now we let the case that $x=a+b+1$ . You can let ${ x=a+b+n }$ where n is natural number, but for simplicity sake we put 1.

This becomes ${ (a+b+1) }^{ 2 }-{ (a+b) }^{ 2 }=ab$ .

After simplification, $2a+2b+1=ab$

$ab-2a-2b-1=0\\ (a-2)(b-2)=5$

$a-2=5\qquad \qquad or\qquad \qquad a-2=1\\ b-2=1\qquad \qquad \qquad \qquad \quad b-2=5$

Both cases give $a+b=10$

So $a+b\ge 10$

This problem presents an interesting computational number theory challenge as an alternative solution. That is, how can you 'creep up' from below to find a minimum when two variables are involved. In this case, since we are interested only in a+b we can consider points in a lattice that is open to infinity and check for whether candidate values of the function are perfect squares using the Babylonian algorithm. It's the for-loop that walks through points in the lattice.

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from itertools import count

# from:
# http://stackoverflow.com/questions/2489435/how-could-i-check-if-a-number-is-a-perfect-square
def is_square(apositiveint):
    x = apositiveint // 2
    seen = set([x])
    while x * x != apositiveint:
        x = (x + (apositiveint // x)) // 2
        if x in seen: return False
        seen.add(x)
    return True

for a,b in ((a,b) for a in count(1) for b in range(1,a)):
    if is_square(a**2+3*a*b+b**2):
        print (a+b)
        break

Taking a^2 + 3ab + b^2 = n^2, one can write (a+b)^2 + ab = n^2, or ab = [n+(a+b)][n-(a+b)] by the difference of two squares. Now set:

a = n + a + b (i)

b = n - a - b (ii)

which adding (i) with (ii) yields a + b = 2n (for positive integers n).

Now, we check ordered pairs (a,b) with distinct positive integers a < b and with trinomial value T = a^2 + 3ab + b^2:

n = 4: (a,b) = (1,3) => T = 19.

n = 6: (a,b) = (1,5) => T = 41; (a,b) = (2,4) => T = 44.

n = 8: (a,b) = (1,7) => T = 71; (a,b) = (2,6) => T = 76; (a,b) = (3,5) => T = 79.

n = 10: (a,b) = (1,9) => T = 109; (a,b) = (2,8) => T = 116; (a,b) = (3,7) => T = 121; (a,b) = (4,6) => T = 124.

The first and smallest perfect square occurs at (a,b) = (3,7), hence a + b = 10 is the smallest possible value under these conditions.