Greedy Calvinosaurus Dinosaur

$\gcd(122 + n^2,122 + (n + 1)^2)$

$= \gcd(122 + n^2,2n + 1)$

Since 2n + 1 is odd, multiplying $122 + n^2$ with 2 will not change the gcd.

$= \gcd(244 + 2n^2,2n + 1)$

Dividing $244 + 2n^2$ by $2n + 1$

$= \gcd(244 - n,2n + 1)$

Again, since $2n + 1$ is odd, multiplying 2n + 1 with 2 will not affect the gcd.

$= \gcd(488 - 2n,2n + 1)$

Dividing 488 - 2n by 2n+1

$= \gcd(489, 2n + 1)$

Therefore largest value of gcd will be 489, when n is a multiple of 489

We let $d_n$ be the greatest common divisor of $122+n^2$ and $122+(n+1)^2$ .

By Euclidean Algorithm that $d_n$ is equal to $(122+(n+1)^2)-(122+n^2)=2n+1$ . $d_n$ is a factor of $(2n+1)^2=4n^2+4n+1$ . Note that it is also a factor of $4(122+n^2)=488+4n^2$ . Hence it is a factor of their difference, which is $4n-487$ . Also, $d_n$ divides $2(2n+1)=4n+2$ . Since it divides $4n-487$ and $4n+2$ , it must divide their difference, which is $489$ .

Now, we would show that $489$ is achievable.

We take $n=244$ . $122+244^2=122(1+2(244))=122(489)$ , and $2(244)+1=489$ , which is the difference of $122+244^2$ and $122+245^2$ .

Hence $489$ is achievable, and the answer is $\boxed{489}$ .

We will define [a,b] as the gcd of the numbers a,b for convenience. Using the Euclidean Algorithm, it is clear that $[122+n^2, 122+n^2+2n+1]=[122+n^2, 2n+1]$ . However, $2n+1$ will never be a multiple of 4, so we can multiply the expression $122+n^2$ , as it won't affect the gcd. Hence, we get $[n^2+122, 2n+1]=[4n^2+488, 2n+1]$ Dividing the latter expression from the first we get $2n-1+ \frac{489}{2n+1}$ . This means that when subtracting (hint, hint, euclidean algorithm) the second expression from the first expression $2n-1$ times, an integer, we have a remainder of 489. Hence this is the equivalence of euclidean algorithm, so we now have that $[122+n^2, 122+(n+1)^2]=[2n+1, 489]$ . The prime factorization of 489 is $3 \cdot 163$ , so if $2n+1<489$ either $2n+1=1,3,163,489$ . Of course we wish to maximize the gcd of both of these expressions/numbers, and if $n=244$ , the gcd could be 489. Obviously, if $2n+1>489$ , then we would maximize it's gcd if the gcd was 489, and this occurs for $n=489k+244$ , for positive integers $k$ . Hence, our gcd is $\boxed{489}$ .

We know that, $(a,b) = (a\pm b)$

so, $(122 + n^{2},122+ (n+1))^{2})$ $= (122 + n^{2},122+(n+1)^{2} - 122+n^{2})$ $= (122 + n^{2},1+2n)$

Note that 1+2n can never be even. so we can multiply 2 in its complementry term without affecting their GCD.

$= 2*(122 + n^{2},1+2n)$ $= (122 + n^{2} - n(1+2n),1+2n)$ $= (244 -n,1+2n)$ $= (2*(244 -n),1+2n)$ $= (488-2n + 1+2n,1+2n)$ $= (489,1+2n)$

So the req. GCD must divide 489 and 2n+1 and so higesht such GCD would be 489.

We will make use of the following known result: If $a,b,q,$ and $r$ are integers with $a=bq+r$ then $\gcd(a,b) = \gcd(b,r).$

In our problem, we have $122+(n+1)^2 = (122+n^2) + (2n+1)$ and $4(122+n^2) = (2n-1)(2n+1)+489.$ Therefore we have $\gcd(122+(n+1)^2,122+n^2) = \gcd(122+n^2, 2n+1)$ and $\gcd(4(122+n^2), 2n+1) = \gcd(2n+1,489).$ Since $2n+1$ is odd, it follows that $\gcd(4(122+n^2), 2n+1) = \gcd(122+n^2, 2n+1).$ Hence $\gcd(122+(n+1)^2,122+n^2) = \gcd(2n+1,489),$ the greatest value of which is $\boxed{489}$ when $2n+1 = 489$ or when $n=244.$