Find a square root of two modulo forty-nine, starting with the root three modulo seven. Testing every integer throws away information: every possible lift of this particular root is 3+7t, with t one of seven residues. The task is to determine one new base-seven digit. A different initial root, four modulo seven, starts a different branch. We lift a specified branch rather than claiming that one computation lists every root.
Predict
Predict whether exactly one digit should work. Expanding gives (3+7t)²−2=7+42t+49t². Divisibility by forty-nine is equivalent to 1+6t=0 modulo seven, so t=1 and the lifted root is ten. Check 100−2=98. The coefficient six is invertible modulo seven; that invertibility, rather than the visual smallness of the correction, explains uniqueness.
Experiment
Complete a second row by hand before reading its result. From root ten modulo forty-nine, write the next root as 10+49t. Divide the constant residual 98 by forty-nine, obtaining two; the derivative twenty reduces to six modulo seven. Thus 2+6t=0 gives t=2 and root 108 modulo 343.
Stage
Root
Residual
Modulus
Initial
3
7
7
First lift
10
98
49
Second lift
108
11662
343
Verify each residual is divisible by its displayed modulus. The residual need not be small as an ordinary real number.
Understand
Let f be a polynomial with integer coefficients, p a prime, k≥1, and p^k divide f(r). Every lift of r to modulus p^(k+1) has the form r+t p^k, with t modulo p. Binomial expansion gives f(r+t p^k) congruent to f(r)+t p^k f′(r) modulo p^(k+1): all terms of degree at least two in the increment contain p^(2k), and 2k≥k+1. After division by p^k the lifting condition is f(r)/p^k+t f′(r)=0 modulo p. If the derivative is nonzero modulo p, multiplication by its inverse gives one digit. If the derivative is zero, the constant term decides: zero permits all p digits; nonzero permits none. This proves the basic lifting theorem and its singular alternatives.
Apply
Diagnose an incorrect universal claim that every modular root lifts uniquely. For f(x)=x² and root zero modulo p, all p choices tp lift modulo p². For f(x)=x²−p, the same starting residue has no lift. A useful implementation therefore stores the modulus, root and residual quotient, checks the derivative, and branches explicitly. Replacing a singular derivative by a floating-point reciprocal destroys the arithmetic argument. Integer polynomial evaluation and exact divisibility checks suffice for this bounded task.
Continue
The repeated unique lifts are compatible residues: reducing a later root returns the earlier root. This introduces the idea of p-adic approximation without identifying it with real convergence. The roots ten and 108 are farther apart on the real line while agreeing to more base-seven digits. General p-adic existence and completeness require an additional development. Here the proved conclusion is exact existence and uniqueness at each finite prime-power stage under the simple-root condition.
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Lift the root 2 of x²−4 modulo 5 to 2+5t modulo 25. Find t in 0,…, 4.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
Lift root 1 of x²−6 modulo 5 to a root in 0,…, 24.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
Prove uniqueness of a lift when p does not divide f′(r).
Structure your argument and preserve revisions
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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Show the root 0 of x² modulo p has p lifts modulo p².
Structure your argument and preserve revisions
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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Show the root 0 of x²−p modulo p has no lift modulo p².
Structure your argument and preserve revisions
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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Explain why a simple root stays simple after lifting.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Reflect and continue
Before marking this lesson complete, explain one key idea without the worked solution and identify a condition that matters. Marking complete records participation, not independent mastery.
