1) You didn’t say integer solutions only. 2) because m is an integer doesn’t force sqrt(m) to be an integer and hence u didn’t have to be an integer. I sure you started with your solution and worked back to create your question… you should ask your question with all your little bits of extra information required to get to the solution you want.
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49-9=40
5:44 (x+y) (x-y)=20x2 has integer solutions, 11 and 9.
but when x and y got substituted, u is not an integer
1) You didn’t say integer solutions only. 2) because m is an integer doesn’t force sqrt(m) to be an integer and hence u didn’t have to be an integer.
I sure you started with your solution and worked back to create your question… you should ask your question with all your little bits of extra information required to get to the solution you want.
Happy New year 2025 to you, dear teacher. I thank you for all your efforts to prepare beautiful math problems welcomed by most of your followers.
Thanks for the kind words, Happy New Year 2025! 🥂🎉🥳
To verify, you don't equate the LHS to 40 ,you just world out what the LHS is and if it comes to 40 the solution(s) are verified
No such thing as a Stanford university admission test. And if there was, this question would never qualify as it is way too easy.
let u^2=(V7)^(Vm) , & v^2=(V3)^(Vm) , I count on one occasion , u^2-v^2=10*4 , (u+v)(u-v)=10*4 , let u+v=10 & u-v=4 ,
u+v+u-v=14 , 2u=14 , u=7 , u^2=49 , (V7)^(Vm)=49 , Vm=2*log7/((log7)/2) , Vm=4 , m=16 ,
test , (V7)^(V16)-(V3)^(V16)=(V7)^4-(V3)^4 , 49-9=40 , OK , solu , m=16 ,
Square root of m can be written m^1/2 not m/2.
m=16 by inspection. No algebra needed. At this level, should it even exist, it’s obvious 49-9 is required.
This can be proven in 30 seconds so why take 11 minutes
mó2