If 4t^2-r^2=17 for positive integers t and r, what is the value of t+r?
A)8
B)11
C)13
D)14
E)15
No solutions in integers.
Taking the equation mod 4, we have that -r^2 ≡ 1 (mod 4), or r^2 ≡ -1 ≡ 3 (mod 4). But perfect squares only leave a remainder of 0 or 1, not 3.
Alternatively, difference of squares gives us
(2t-r)(2t+r) = 17
2t+r is a positive integer, so 2t-r is also a positive integer. The only factorization of 17 in positive integers is 1*17, and since 2t-r < 2t+r, we have 2t-r = 1, 2t+r = 17. Solving gives t = 9/2, r = 8, which is not an integer solution.
Did you typo somewhere?
It is written this way in the book 
I’m 100% sure real test 
Then the problem is either flawed or you typed it incorrectly but didn’t notice – clearly there are no integer solutions.
I took that test not sure what month. The question was like this
9t^2-r^2=17
Here is the trick
(3t+r)(3t-r)=17
You know that 17 is prime so you need (17)(1). T=3 and R=8
T+R=11
^Yeah, if 4 is replaced with 9, then it’s solvable using the method @Jr12317 and I showed above.
Yeah it’s 9 actually!!! Thanks a lotttttttt guys!!! GREAT JOB!!