Solving a Russian Math Olympiad Problem | Solve for x If 4^27 + 4^1000 + 4^x is a Perfect Square

🔴Solving a Russian Math Olympiad Problem | Solve for the Largest x for Which 4^27 + 4^1000 + 4^x would be a Perfect Square |

Hey there.

Today, we are dealing with a nice Russian Math Olympiad problem, in which we want to solve for the largest integer x, such that 4^27 + 4^1000 + 4^x would be a perfect square. Now finding an integer that will make 4^27 + 4^1000 + 4^x a perfect square, is not really hard, but the tricky part of this question is that we don't know if the x we find is the largest x possible or not.
So in order to solve this interesting Russian Math Olympiad question, we will find a value for x, and then we will try to find out if there is a larger value for x or not.

🔴I hope you enjoy watching this video on a really nice Russian Math Olympiad problem .🔴

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topics covered in this video:
Prove that this expression is a perfect square
solve for x
solving a nice math olympiad problem
solving a Russian math olympiad problem
how to solve for x if 4^27 + 4^1000 + 4^x is a perfect square

#SolveForX #RussianMathOlympiad #matholympiad #AnonMath #QuadraticEquation #matholympiadquestion

[Homcomru]

Honestly the craziest thing is that I swear this question not only appears in Russian maths Olympiads, but also in the Russian State Exam.

Makes me really remember how Russia is quite advanced in terms of the math curriculum.

[SuperYoonHo]

Thank you I just love exponents💯💯🙏🙏👍👍✅✅

[georget]

Please review and comment on the following:

4^27+4^1000+4^× = y^2
The left side is even.
Thus, y has to be even too.

I move the 4^x to the right side and I transform the right side into a difference of squares.

4^27 + 4^1000 = (y - 2^x)*(y + 2^x)
4^27 * (1 + 4^973) = (y - 2^x)*(y + 2^x)

The left side is odd.
The right side is even.

Thus, the initial equation cannot be a perfect square.